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Permutation implies arrangement where order of things is important and includes word formation number formation, circular permutation etc.
- All permutations (or arrangements) made with the letters a, b, c by taking two at a time are (ab, ba, ac, ca, bc, cb).
- All permutations made with the letters a, b, c taking all at a time are (abc, acb, bac, bca, cab, cba).
Number of Permutations:
Number of all permutations of n things, taken r at a time, is given by:
nPr = n(n – 1)(n – 2) … (n – r + 1) = n! / (n – r)!
Examples:
- 6P2 = (6 x 5) = 30.
- 7P3 = (7 x 6 x 5) = 210.
More information: https://en.wikipedia.org/wiki/Permutation
Combination means selection where order is not important and it involves selection of team forming geometrical figures, distribution of things etc.
Examples:
- Suppose we want to select two out of three boys A, B, C. Then, possible selections are AB, BC and CA. [Note: AB and BA represent the same selection]
- All the combinations formed by a, b, c taking ab, bc, ca.
- The only combination that can be formed of three letters a, b, c taken all at a time is abc.
- Various groups of 2 out of four persons A, B, C, D are: AB, AC, AD, BC, BD, CD.
- Note that ab ba are two different permutations but they represent the same combination.
Number of Combinations:
The number of all combinations of n things, taken r at a time is:
nCr = n ! / [(r!)(n-r)]! = [n(n-1)(n-2)…. To r factors] / r!
Note: nCn = 1 and nC0 = 1 , nCr = nC(n – r)
Examples:
- 11C4 = (11 x 10 x 9 x 8) / (4 x 3 x 2 x 1) = 330
- 16C13 = 16C(16 – 13) = 16C3 = 16 x 15 x 14 / 3!= (16 x 15 x 14) / (3 x 2 x 1) = 560
Factorial Notation
Let n be a positive integer. Then, factorial n, denoted n! is defined as:
n! = n(n – 1)(n – 2) … 3.2.1.
Examples:
- We define 0! = 1.
- 4! = (4 x 3 x 2 x 1) = 24.
- 5! = (5 x 4 x 3 x 2 x 1) = 120.
More information: https://en.wikipedia.org/wiki/Combination
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Permutation and combination - Question and Answers
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Question 1 |
In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
266 | |
5040 | |
11760 | |
86400 |
Question 1 Explanation:
Required number of ways:
= (8C5 X 10C6)
= {(8*7*6/3*2*1)*(10*9*8*7/4*3*2*1)}
= 11760
= (8C5 X 10C6)
= {(8*7*6/3*2*1)*(10*9*8*7/4*3*2*1)}
= 11760
Question 2 |
How many ways to draw 3balls from 16 balls pool.(balls are numbered as 1 to 16 and are distinct)
16P3 | |
16C3 | |
16! | |
72 |
Question 2 Explanation:
No. of ways to draw 3 balls from 16 balls: 16C3
Question 3 |
From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
564 | |
645 | |
735 | |
756 |
Question 3 Explanation:
We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
Required number of ways = (7C3 x 6C2) + (7C4 x 6C1) + (7C5)
(7 x 6 x 5)/(3x2x1) x (6 x 5/2x1) + (7C3 x 6C1) + (7C2)
=525 +(7x6x5/3x2x1 x 6) + (7x6/2x1)
= (525 + 210 + 21)
=756
(7 x 6 x 5)/(3x2x1) x (6 x 5/2x1) + (7C3 x 6C1) + (7C2)
=525 +(7x6x5/3x2x1 x 6) + (7x6/2x1)
= (525 + 210 + 21)
=756
Question 4 |
In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
360 | |
480 | |
720 | |
5040 |
Question 4 Explanation:
The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Required number of ways = (120 x 6) = 720.
Question 5 |
In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?
810 | |
1440 | |
2880 | |
50400 |
Question 5 Explanation:
In the word 'CORPORATION', we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters = 7!/2! = 2520.
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged
in 5!/3! = 20 ways.
Required number of ways = (2520 x 20) = 50400.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters = 7!/2! = 2520.
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged
in 5!/3! = 20 ways.
Required number of ways = (2520 x 20) = 50400.
Question 6 |
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
210 | |
1050 | |
25200 | |
21400 |
Question 6 Explanation:
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
(7C3 x 4C2)
(7*6*5/3*2*1) * (4*3/2*1)
=210
Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
Number of ways of arranging
5 letters among themselves = 5!
= 5 x 4 x 3 x 2 x 1
= 120.
Required number of ways = (210 x 120) = 25200.
(7C3 x 4C2)
(7*6*5/3*2*1) * (4*3/2*1)
=210
Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
Number of ways of arranging
5 letters among themselves = 5!
= 5 x 4 x 3 x 2 x 1
= 120.
Required number of ways = (210 x 120) = 25200.
Question 7 |
In how many ways can the letters of the word 'LEADER' be arranged?
72 | |
144 | |
360 | |
720 |
Question 7 Explanation:
The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
Required number of ways =6!/(1!)(2!)(1!)(1!)(1!)
= 360.
Required number of ways =6!/(1!)(2!)(1!)(1!)(1!)
= 360.
Question 8 |
In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
63 | |
90 | |
126 | |
45 |
Question 8 Explanation:
Required number of ways = (7C5 x 3C2) =
(7C2 x 3C1) =
(7*6/2*3) = 63
(7C2 x 3C1) =
(7*6/2*3) = 63
Question 9 |
How many 4-letter words with or without meaning, can be formed out of the letters of the word, 'LOGARITHMS', if repetition of letters is not allowed?
40 | |
400 | |
5040 | |
2520 |
Question 9 Explanation:
'LOGARITHMS' contains 10 different letters.
Required number of words = Number of arrangements of 10 letters, taking 4 at a time..
= 10P4.
= (10 x 9 x 8 x 7).
= 5040.
Required number of words = Number of arrangements of 10 letters, taking 4 at a time..
= 10P4.
= (10 x 9 x 8 x 7).
= 5040.
Question 10 |
In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together?
10080 | |
4989600 | |
120960 | |
None of these |
Question 10 Explanation:
In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
Number of ways of arranging these letters = 8!/(2!)(2!) = 10080.
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters = 4!/2! = 12..
Required number of words = (10080 x 12) = 120960.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
Number of ways of arranging these letters = 8!/(2!)(2!) = 10080.
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters = 4!/2! = 12..
Required number of words = (10080 x 12) = 120960.
Question 11 |
In how many different ways can the letters of the word 'OPTICAL' be arranged so that the vowels always come together?
120 | |
720 | |
4320 | |
2160 |
Question 11 Explanation:
The word 'OPTICAL' contains 7 different letters.
When the vowels OIA are always together, they can be supposed to form one letter..
Then, we have to arrange the letters PTCL (OIA)..
Now, 5 letters can be arranged in 5! = 120 ways..
The vowels (OIA) can be arranged among themselves in 3! = 6 ways..
Required number of ways = (120 x 6) = 720
When the vowels OIA are always together, they can be supposed to form one letter..
Then, we have to arrange the letters PTCL (OIA)..
Now, 5 letters can be arranged in 5! = 120 ways..
The vowels (OIA) can be arranged among themselves in 3! = 6 ways..
Required number of ways = (120 x 6) = 720
Question 12 |
In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
159 | |
194 | |
205 | |
209 |
Question 12 Explanation:
We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
number of ways = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
=(6*4) +((6*5/2*1) *(4*3/2*1)) + ((6*5*4/3*2*1)*4)+(6*5/2*1)
= (24 + 90 + 80 + 15)
= 209.
number of ways = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
=(6*4) +((6*5/2*1) *(4*3/2*1)) + ((6*5*4/3*2*1)*4)+(6*5/2*1)
= (24 + 90 + 80 + 15)
= 209.
Question 13 |
How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated?
5 | |
10 | |
15 | |
20 |
Question 13 Explanation:
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20.
Question 14 |
In how many ways a committee, consisting of 5 men and 6 women can be formed from 8 men and 10 women?
266 | |
5040 | |
11760 | |
86400 |
Question 14 Explanation:
Required number of ways = (8C5 x 10C6)
(8C3 x 10C4)
= (8 x 7 x 6 x/3 x 2 x 1) x (10 x 9 x 8 x 7 /4 x 3 x 2 x 1)
=11760 = 11760.
(8C3 x 10C4)
= (8 x 7 x 6 x/3 x 2 x 1) x (10 x 9 x 8 x 7 /4 x 3 x 2 x 1)
=11760 = 11760.
Question 15 |
A box contains 2 white balls, 3 black balls and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least one black ball is to be included in the draw?
32 | |
48 | |
64 | |
96 |
Question 15 Explanation:
We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
Required number of ways = (3C1 x 6C2) + (3C2 x 6C1) + (3C3)
=(3* 6x5/2x1) + (3x2/2x1 * 6) + 1
= (45 + 18 + 1)
= 64.
Required number of ways = (3C1 x 6C2) + (3C2 x 6C1) + (3C3)
=(3* 6x5/2x1) + (3x2/2x1 * 6) + 1
= (45 + 18 + 1)
= 64.
Question 16 |
In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?
32 | |
48 | |
36 | |
60 |
Question 16 Explanation:
There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.
Let us mark these positions as under: (1) (2) (3) (4) (5) (6)
Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.
Number of ways of arranging the vowels = 3P3 = 3! = 6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = 3P3 = 3! = 6.
Total number of ways = (6 x 6) = 36
Let us mark these positions as under: (1) (2) (3) (4) (5) (6)
Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.
Number of ways of arranging the vowels = 3P3 = 3! = 6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = 3P3 = 3! = 6.
Total number of ways = (6 x 6) = 36
Question 17 |
How many multiples of 5 are there from 10 to 95?
20 | |
18 | |
17 | |
25 |
Question 17 Explanation:
As you know, multiples of 5 are integers having 0 or 5 in the digit to the extreme right (i.e. the unit's place).
The first digit from the right can be chosen in 2 ways.
The second digit can be any one of 1,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,9
i.e. There are 9 choices for the second digit.
Thus, there are 2×9=2×9= 18 multiples of 5 from 10 to 95.
The second digit can be any one of 1,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,9
i.e. There are 9 choices for the second digit.
Thus, there are 2×9=2×9= 18 multiples of 5 from 10 to 95.
Question 18 |
In a city, the bus route numbers consist of a natural number less than 100, followed by one of the letters A, B, C, D, E and F. How many different bus routes are possible?
456 | |
522 | |
594 | |
500 |
Question 18 Explanation:
The number can be any one of the natural numbers from 1 to 99.
There are 99 choices for the number.
The letter can be chosen in 6 ways.
Number of possible bus routes are 99×6=99×6= 594
There are 99 choices for the number.
The letter can be chosen in 6 ways.
Number of possible bus routes are 99×6=99×6= 594
Question 19 |
There are 3 questions in a question paper. If the questions have 4, 3 and 2 solutions respectively, find the total number of solutions.
20 | |
15 | |
45 | |
24 |
Question 19 Explanation:
Here question 1 has 4 solutions, question 2 has 3 solutions and question 3 has 2 solutions.
=> By the multiplication (counting) rule,
Total number of solutions=4×3×2==4×3×2= 24
=> By the multiplication (counting) rule,
Total number of solutions=4×3×2==4×3×2= 24
Question 20 |
Consider the word ROTOR. Whichever way you read it, from left to right or from right to left, you get the same word. Such a word is known as palindrome. Find the maximum possible number of 5-letter palindromes.
17576 | |
17654 | |
14789 | |
16872 |
Question 20 Explanation:
The first letter from the right can be chosen in 26 ways because there are 26 alphabets.
Having chosen this, the second letter can be chosen in 26 ways.
=> The first two letters can be chosen in 26×26=67626×26=676 ways
Having chosen the first two letters, the third letter can be chosen in 26 ways.
=> All the three letters can be chosen in 676×26=17576676×26=17576 ways.
It implies that the maximum possible number of five letter palindromes is 17576 because the fourth letter is the same as the second letter and the fifth letter is the same as the first letter.
Having chosen this, the second letter can be chosen in 26 ways.
=> The first two letters can be chosen in 26×26=67626×26=676 ways
Having chosen the first two letters, the third letter can be chosen in 26 ways.
=> All the three letters can be chosen in 676×26=17576676×26=17576 ways.
It implies that the maximum possible number of five letter palindromes is 17576 because the fourth letter is the same as the second letter and the fifth letter is the same as the first letter.
Question 21 |
How many 3-digit numbers can be formed with the digits 1,4,7,8 and 9 if the digits are not repeated?
30 | |
40 | |
60 | |
59 |
Question 21 Explanation:
Three digit numbers will have unit's, ten's and hundred's place.
Out of 5 given digits any one can take the unit's place. This can be done in 5 ways ------- (i).
After filling the unit's place, any of the four remaining digits can take the ten's place..
This can be done in 4 ways ------- (ii).
After filling in ten's place, hundred's place can be filled from any of the three remaining digits..
This can be done in 3 ways ------- (iii).
=> By counting principle, the number of 3 digit numbers =5×4×3==5×4×3= 60
Out of 5 given digits any one can take the unit's place. This can be done in 5 ways ------- (i).
After filling the unit's place, any of the four remaining digits can take the ten's place..
This can be done in 4 ways ------- (ii).
After filling in ten's place, hundred's place can be filled from any of the three remaining digits..
This can be done in 3 ways ------- (iii).
=> By counting principle, the number of 3 digit numbers =5×4×3==5×4×3= 60
Question 22 |
Suppose you can travel from a place A to a place B by 3 buses, from place B to place C by 4 buses, from place C to place D by 2 buses and from place D to place E by 3 buses. In how many ways can you travel from A to E?
63 | |
45 | |
72 | |
85 |
Question 22 Explanation:
The bus from A to B can be selected in 3 ways.
The bus from B to C can be selected in 4 ways.
The bus from C to D can be selected in 2 ways.
The bus from D to E can be selected in 3 ways.
So, by the General Counting Principle, one can travel from A to E in 3×4×2×3=3×4×2×3= 72 ways
The bus from B to C can be selected in 4 ways.
The bus from C to D can be selected in 2 ways.
The bus from D to E can be selected in 3 ways.
So, by the General Counting Principle, one can travel from A to E in 3×4×2×3=3×4×2×3= 72 ways
Question 23 |
Suppose you want to arrange your English, Hindi, Mathematics, History, Geography and Science books on a shelf. In how many ways can you do it?
630 | |
569 | |
654 | |
720 |
Question 23 Explanation:
We have to arrange 6 books.
The number of permutations of n objects isn!=n.(n−1).(n−2)...2.1n!=n.(n−1).(n−2)...2.1.
Here n=6n=6 and therefore, number of permutations is 6.5.4.3.2.1=6.5.4.3.2.1= 720.
The number of permutations of n objects isn!=n.(n−1).(n−2)...2.1n!=n.(n−1).(n−2)...2.1.
Here n=6n=6 and therefore, number of permutations is 6.5.4.3.2.1=6.5.4.3.2.1= 720.
Question 24 |
If you have 6 New Year greeting cards and you want to send them to 4 of your friends, in how many ways can this are done?
260 | |
392 | |
723 | |
360 |
Question 24 Explanation:
We have to find number of permutations of 4 objects out of 6 objects.
This number is 6P4=6(6−1)(6−2)
(6−3)=6×5×4×3=3606P4=6(6−1)(6−2)
(6−3)=6×5×4×3=360
This number is 6P4=6(6−1)(6−2)
(6−3)=6×5×4×3=3606P4=6(6−1)(6−2)
(6−3)=6×5×4×3=360
Question 25 |
Suppose 7 students are staying in a hall in a hostel and they are allotted 7 beds. Among them, Parvin does not want a bed next to Anju because Anju snores. Then, in how many ways can you allot the beds?
200ways | |
5400ways | |
3600ways | |
3505ways |
Question 25 Explanation:
Let the beds be numbered 1 to 7.
Case 1:
Suppose Anju is allotted bed number 1.
Then, Parvin cannot be allotted bed number 2.
So Parvin can be allotted a bed in 5 ways.
After allotting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways.
So, in this case the beds can be allotted in 5×5!=6005×5!=600 ways.
Case 2:
Anju is allotted bed number 7.
Then, Parvin cannot be allotted bed number 6
As in Case 1, the beds can be allotted in 600 ways.
Case 3:
Anju is allotted one of the beds numbered 2,3,4,52,3,4,5 or 66
Parvin cannot be allotted the beds on the right hand side and left hand side of Anju's bed.
For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin.
Therefore, Parvin can be allotted a bed in 4 ways in all these cases.
After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways.
Therefore, in each of these cases, the beds can be allotted 4×5!=4804×5!=480 ways.
=> The beds can be allotted in:
2×600+5×480=1200+2400=2×600+5×480=1200+2400= 3600 ways
Case 1:
Suppose Anju is allotted bed number 1.
Then, Parvin cannot be allotted bed number 2.
So Parvin can be allotted a bed in 5 ways.
After allotting a bed to Parvin, the remaining 5 students can be allotted beds in 5! ways.
So, in this case the beds can be allotted in 5×5!=6005×5!=600 ways.
Case 2:
Anju is allotted bed number 7.
Then, Parvin cannot be allotted bed number 6
As in Case 1, the beds can be allotted in 600 ways.
Case 3:
Anju is allotted one of the beds numbered 2,3,4,52,3,4,5 or 66
Parvin cannot be allotted the beds on the right hand side and left hand side of Anju's bed.
For example, if Anju is allotted bed number 2, beds numbered 1 or 3 cannot be allotted to Parvin.
Therefore, Parvin can be allotted a bed in 4 ways in all these cases.
After allotting a bed to Parvin, the other 5 can be allotted a bed in 5! ways.
Therefore, in each of these cases, the beds can be allotted 4×5!=4804×5!=480 ways.
=> The beds can be allotted in:
2×600+5×480=1200+2400=2×600+5×480=1200+2400= 3600 ways
Question 26 |
There are 4 books on fairy tales, 5 novels and 3 plays. In how many ways can you arrange these so that books on fairy tales are together, novels are together and plays are together and in the order, books on fairy tales, novels and plays.
28756 | |
28756 | |
17280 | |
16547 |
Question 26 Explanation:
There are 4 books on fairy tales and they have to be put together.
They can be arranged in 4! ways.
Similarly, there are 5 novels.
They can be arranged in 5! ways.
And there are 3 plays.
They can be arranged in 3! ways.
So, by the counting principle all of them together can be arranged in 4!×5!×3!=4!×5!×3!= 17280 ways
They can be arranged in 4! ways.
Similarly, there are 5 novels.
They can be arranged in 5! ways.
And there are 3 plays.
They can be arranged in 3! ways.
So, by the counting principle all of them together can be arranged in 4!×5!×3!=4!×5!×3!= 17280 ways
Question 27 |
Suppose there are 4 books on fairy tales, 5 novels and 3 plays as in Example 5.3. They have to be arranged so that the books on fairy tales are together, novels are together and plays are together, but we no longer require that they should be in a specific order. In how many ways can this be done?
465765 | |
678954 | |
103680 | |
876524 |
Question 27 Explanation:
First, we consider the books on fairy tales, novels and plays as single objects.
These three objects can be arranged in 3!=63!=6 ways.
Let us fix one of these 6 arrangements.
This may give us a specific order, say, novels -> fairy tales -> plays.
Given this order, the books on the same subject can be arranged as follows.
The 4 books on fairy tales can be arranged among themselves in 4!=244!=24 ways.
The 5 novels can be arranged in 5!=1205!=120 ways.
The 3 plays can be arranged in 3!=63!=6 ways.
For a given order, the books can be arranged in 24×120×6=1728024×120×6=17280 ways.
Therefore, for all the 6 possible orders the books can be arranged in 6×17280=6×17280= 103680 ways.
These three objects can be arranged in 3!=63!=6 ways.
Let us fix one of these 6 arrangements.
This may give us a specific order, say, novels -> fairy tales -> plays.
Given this order, the books on the same subject can be arranged as follows.
The 4 books on fairy tales can be arranged among themselves in 4!=244!=24 ways.
The 5 novels can be arranged in 5!=1205!=120 ways.
The 3 plays can be arranged in 3!=63!=6 ways.
For a given order, the books can be arranged in 24×120×6=1728024×120×6=17280 ways.
Therefore, for all the 6 possible orders the books can be arranged in 6×17280=6×17280= 103680 ways.
Question 28 |
In how many ways can 4 girls and 5 boys be arranged in a row so that all the four girls are together?
17654 | |
17280 | |
56783 | |
78345 |
Question 28 Explanation:
Let 4 girls be one unit and now there are 6 units in all.
They can be arranged in 6! ways.
In each of these arrangements 4 girls can be arranged in 4! ways.
=> Total number of arrangements in which girls are always together
=6!×4!=720×24==6!×4!=720×24= 17280
In each of these arrangements 4 girls can be arranged in 4! ways.
=> Total number of arrangements in which girls are always together
=6!×4!=720×24==6!×4!=720×24= 17280
Question 29 |
Find the number of subsets of the set {1,2,3,4,5,6,7,8,9,10,11}{1,2,3,4,5,6,7,8,9,10,11} having 4 elements.
330 ways | |
340 ways | |
333 ways | |
348 ways |
Question 29 Explanation:
Here the order of choosing the elements doesn't matter and this is a problem in combinations.
We have to find the number of ways of choosing 4 elements of this set which has 11 elements.
This can be done in:
11C4=11×10×9×81×2×3×4=11C4=11×10×9×81×2×3×4= 330 ways
We have to find the number of ways of choosing 4 elements of this set which has 11 elements.
This can be done in:
11C4=11×10×9×81×2×3×4=11C4=11×10×9×81×2×3×4= 330 ways
Question 30 |
In a box, there are 5 black pens, 3 white pens and 4 red pens. In how many ways can 2 black pens, 2 white pens and 2 red pens can be chosen?
240ways | |
450ways | |
180ways | |
220ways |
Question 30 Explanation:
Number of ways of choosing 2 black pens from 5 black pens
=5C2=5P22!=5×41×2=10=5C2=5P22!=5×41×2=10
Number of ways of choosing 2 white pens from 3 white pens
=3C2=3P22!=3×21×2=3=3C2=3P22!=3×21×2=3
Number of ways of choosing 2 red pens from 4 red pens
=4C2=4P22!=4×31×2=6=4C2=4P22!=4×31×2=6
=> By the Counting Principle, 2 black pens, 2 white pens, and 2 red pens can be chosen in
10×3×6=10×3×6= 180ways.
=5C2=5P22!=5×41×2=10=5C2=5P22!=5×41×2=10
Number of ways of choosing 2 white pens from 3 white pens
=3C2=3P22!=3×21×2=3=3C2=3P22!=3×21×2=3
Number of ways of choosing 2 red pens from 4 red pens
=4C2=4P22!=4×31×2=6=4C2=4P22!=4×31×2=6
=> By the Counting Principle, 2 black pens, 2 white pens, and 2 red pens can be chosen in
10×3×6=10×3×6= 180ways.
Question 31 |
There are 5 novels and 4 biographies. In how many ways can 4 novels and 2 biographies can be arranged on a shelf?
11234 | |
21600 | |
37432 | |
37432 |
Question 31 Explanation:
4 novels can be selected out of 5 in 5C45C4 ways.
2 biographies can be selected out of4 in 4C24C2 ways.
Number of ways of arranging novels and biographies: =5C4×4C2=30=5C4×4C2=30
After selecting any 6 books (4 novels and 2 biographies) in one of the 30 ways, they can be arranged on the shelf in 6!=7206!=720 ways.
By the Counting Principle, the total number of arrangements =30×720==30×720= 21600
2 biographies can be selected out of4 in 4C24C2 ways.
Number of ways of arranging novels and biographies: =5C4×4C2=30=5C4×4C2=30
After selecting any 6 books (4 novels and 2 biographies) in one of the 30 ways, they can be arranged on the shelf in 6!=7206!=720 ways.
By the Counting Principle, the total number of arrangements =30×720==30×720= 21600
Question 32 |
Evaluate 7P2. 4P3
1456 | |
1008 | |
1080 | |
1000 |
Question 32 Explanation:
(7!/ 5!). (4!/ 1!)
⇒(7.6). (4.3.2)
⇒1008
⇒(7.6). (4.3.2)
⇒1008
Question 33 |
Evaluate 5C2. 3C2
20 | |
30 | |
40 | |
60 |
Question 33 Explanation:
(5!/3!2!). (3!/2!1!)
⇒(5.4/2). (3)
⇒30
⇒(5.4/2). (3)
⇒30
Question 34 |
How many ways are there in selecting 5 members from 6 males and 5 females, consisting 3 males and 2 females?
120 | |
240 | |
100 | |
200 |
Question 34 Explanation:
This is a case of combination i.e. selecting 3 males from 6 males and 2 females from 5 females.
⇒Required number of ways = (6C3 *5C2)
⇒(6.5.4/3.2)*(5.4/2)
⇒200.
⇒Required number of ways = (6C3 *5C2)
⇒(6.5.4/3.2)*(5.4/2)
⇒200.
Question 35 |
How many words can be formed by using letters of the word "DAUGHTER" so that the vowels come together?
4320 | |
4563 | |
4632 | |
4700
|
Question 35 Explanation:
This is a case of permutation. In a word "DAUGHTER", there are 8 letters including 3 vowels (AUE)
According to the question, vowels should always come together. Therefore, in this case we will treat all the vowels as one entity or one alphabet. This implies, in total there are 6 words (one word which is a group of vowels)
These 6 words can be arranged in 6P6 ways ⇒6P6 = 6!/1! = 6! = 720 WAYS
Also, three vowels in a group may be arranged in 3! ways
⇒3! = 6 ways
Therefore, required number of words = (720*6) = 4320
According to the question, vowels should always come together. Therefore, in this case we will treat all the vowels as one entity or one alphabet. This implies, in total there are 6 words (one word which is a group of vowels)
These 6 words can be arranged in 6P6 ways ⇒6P6 = 6!/1! = 6! = 720 WAYS
Also, three vowels in a group may be arranged in 3! ways
⇒3! = 6 ways
Therefore, required number of words = (720*6) = 4320
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