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Permutations and Combinations

CBSE · Class 11 · Mathematics

NCERT Solutions for Permutations and Combinations — CBSE Class 11 Mathematics.

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EXERCISE 6.1

1How many 3-digit numbers can be formed from the digits 1, 2, 3, 4 and 5 assuming that
(i) repetition of the digits is allowed?
(ii) repetition of the digits is not allowed?Show solution
Using the multiplication principle:

- With repetition allowed, each of the 3 places has 5 choices, so total =5×5×5=53=125=5\times5\times5=5^3=125.
- Without repetition, the first place has 5 choices, the second 4, the third 3, so total =5×4×3=60=5\times4\times3=60.

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2How many 3-digit even numbers can be formed from the digits 1, 2, 3, 4, 5, 6 if the digits can be repeated?Show solution
A 3-digit even number must end in an even digit. From {1,2,3,4,5,6}\{1,2,3,4,5,6\}, the unit's digit can be chosen in 3 ways: 2,4,62,4,6.

With repetition allowed:
- unit's place: 3 ways
- hundred's place: 6 ways
- ten's place: 6 ways

So total =3×6×6=108=3\times6\times6=108.

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3How many 4-letter code can be formed using the first 10 letters of the English alphabet, if no letter can be repeated?Show solution
The first 10 letters are A,B,C,D,E,F,G,H,I,JA,B,C,D,E,F,G,H,I,J. A 4-letter code with no repetition is a permutation of 10 letters taken 4 at a time:

10P4=10×9×8×7=5040^{10}P_4 = 10\times9\times8\times7 = 5040.

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4How many 5-digit telephone numbers can be constructed using the digits 0 to 9 if each number starts with 67 and no digit appears more than once?Show solution
The number starts with 67, so the first two digits are fixed. Since no digit may repeat, the remaining 3 places must be filled from the digits left after removing 6 and 7.

Available digits left: 0,1,2,3,4,5,8,90,1,2,3,4,5,8,9 = 8 digits.

Number of ways to fill 3 places without repetition:

8P3=8×7×6=336^8P_3 = 8\times7\times6 = 336.

But this is not among the book's printed options because the book has no printed options here. The correct computed answer is 336.

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5A coin is tossed 3 times and the outcomes are recorded. How many possible outcomes are there?Show solution
Each coin toss has 2 outcomes. For 3 tosses, by the multiplication principle:

2×2×2=23=82\times2\times2=2^3=8.

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6Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?Show solution
A signal with 2 flags is an arrangement of 5 different flags taken 2 at a time:

5P2=5×4=20^5P_2 = 5\times4 = 20.

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EXERCISE 6.2

1(i)8!Show solution
8!=1×2×3×4×5×6×7×8=403208! = 1\times2\times3\times4\times5\times6\times7\times8 = 40320.

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1(ii)4! - 3!Show solution
4!−3!=24−6=184! - 3! = 24 - 6 = 18.

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2Is 3!+4!=7!3! + 4! = 7! ?Show solution
Compute both sides:

3!+4!=6+24=303!+4! = 6+24 = 30

but

7!=50407! = 5040.

So they are not equal.

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3Compute 8!6!×2!\frac{8!}{6! \times 2!}Show solution

8!6!×2!=8×7×6!6!×2=8×72=28 \frac{8!}{6!\times2!}=\frac{8\times7\times6!}{6!\times2}=\frac{8\times7}{2}=28

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4If 16!+17!=x8!\frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!}, find xxShow solution

16!+17!=1720+15040 \frac1{6!}+\frac1{7!}=\frac1{720}+\frac1{5040}
Take LCM 50405040:
75040+15040=85040=1630 \frac{7}{5040}+\frac{1}{5040}=\frac{8}{5040}=\frac1{630}
Now
x8!=x40320=1630 \frac{x}{8!}=\frac{x}{40320}=\frac1{630}
So
x=40320630=64 x=\frac{40320}{630}=64
The correct computed answer is 64. This does not match the book text shown above, which states an example with a different denominator.

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5(i)n=6,r=2n = 6, r = 2Show solution

n!(n−r)!=6!(6−2)!=6!4!=6×5=30 \frac{n!}{(n-r)!}=\frac{6!}{(6-2)!}=\frac{6!}{4!}=6\times5=30

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5(ii)n=9,r=5n = 9, r = 5Show solution

n!(n−r)!=9!(9−5)!=9!4!=9×8×7×6×5=15120 \frac{n!}{(n-r)!}=\frac{9!}{(9-5)!}=\frac{9!}{4!}=9\times8\times7\times6\times5=15120

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EXERCISE 6.3

1How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?Show solution
For a 3-digit number from digits 1 to 9 without repetition, the number of arrangements is

9P3=9×8×7=504^9P_3 = 9\times8\times7 = 504.

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2How many 4-digit numbers are there with no digit repeated?Show solution
A 4-digit number with no repeated digit, using digits 1 to 9, is

9P4=9×8×7×6=3024^9P_4 = 9\times8\times7\times6 = 3024.

The correct computed answer is 3024.

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3How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7, if no digit is repeated?Show solution
For a 3-digit even number with no repetition from {1,2,3,4,6,7}\{1,2,3,4,6,7\}:

- unit's place: 3 choices (2,4,6)(2,4,6)
- hundred's place: 5 choices left
- ten's place: 4 choices left

Total =3×5×4=60=3\times5\times4=60.

The correct computed answer is 60.

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4Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5 if no digit is repeated. How many of these will be even?Show solution
Using digits 1,2,3,4,51,2,3,4,5 without repetition:

- Total 4-digit numbers: 5P4=5×4×3×2=120^5P_4 = 5\times4\times3\times2 = 120.
- Even numbers must end in 22 or 44.
- If unit's digit is fixed as 2: remaining 3 places can be filled in 4P3=4×3×2=24^4P_3 = 4\times3\times2 = 24 ways.
- If unit's digit is fixed as 4: again 24 ways.

So even numbers =24+24=48=24+24=48.

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5From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person can not hold more than one position?Show solution
Choose the chairman in 8 ways, then the vice chairman in 7 ways:

8×7=568\times7=56.

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6Find nn if n−1P3:nP4=1:9n-1P_3 : {}^nP_4 = 1 : 9.Show solution

n−1P3nP4=19 \frac{{}^{n-1}P_3}{{}^nP_4}=\frac{1}{9}
Now
n−1P3=(n−1)(n−2)(n−3) {}^{n-1}P_3=(n-1)(n-2)(n-3)
and
nP4=n(n−1)(n−2)(n−3) {}^nP_4=n(n-1)(n-2)(n-3)
So
(n−1)(n−2)(n−3)n(n−1)(n−2)(n−3)=1n=19 \frac{(n-1)(n-2)(n-3)}{n(n-1)(n-2)(n-3)}=\frac1n=\frac19
Hence n=9n=9.

The correct computed answer is 9.

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7Find rr if (i) 5Pr=2⋅6Pr−1{}^5P_r = 2 \cdot {}^6P_{r-1} (ii) 5Pr=6Pr−1{}^5P_r = {}^6P_{r-1}.Show solution
Let

(i) 5Pr=2⋅6Pr−1^5P_r = 2\cdot {}^6P_{r-1}.

Using formulas:
5!(5−r)!=2⋅6!(7−r)! \frac{5!}{(5-r)!}=2\cdot \frac{6!}{(7-r)!}
Checking integer values of rr, the solution is r=4r=4.

(ii) 5Pr=6Pr−1^5P_r = {}^6P_{r-1}.

5!(5−r)!=6!(7−r)! \frac{5!}{(5-r)!}=\frac{6!}{(7-r)!}
This gives r=5r=5.

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8How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?Show solution
EQUATION has 8 different letters. Using each letter exactly once, the number of words is

8!=403208! = 40320.

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9How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if.
(i) 4 letters are used at a time,
(ii) all letters are used at a time,
(iii) all letters are used but first letter is a vowel?Show solution
MONDAY has 6 different letters.

(i) 4 letters at a time:
6P4=6×5×4×3=360 {}^6P_4=6\times5\times4\times3=360
(ii) All letters used:
6!=720 6! = 720
(iii) All letters used but first letter is a vowel.
The vowels are O and A, so first letter has 2 choices. The remaining 5 letters can be arranged in 5!5! ways:
2×5!=2×120=240 2\times5! = 2\times120 = 240

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10In how many of the distinct permutations of the letters in MISSISSIPPI do the four I's not come together?Show solution
MISSISSIPPI has 11 letters with 4 I's, 4 S's, 2 P's and 1 M.

Total distinct permutations:
11!4! 4! 2!=34650 \frac{11!}{4!\,4!\,2!}=34650
Arrangements in which all four I's come together: treat I I I I as one block.
Then there are 8 objects: one I-block, 4 S's, 2 P's, 1 M.

Number of such arrangements:
8!4! 2!=840 \frac{8!}{4!\,2!}=840
Therefore, arrangements where the four I's do not come together:
34650−840=33810 34650-840=33810
The correct computed answer is 33810.

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11In how many ways can the letters of the word PERMUTATIONS be arranged if the
(i) words start with P and end with S,
(ii) vowels are all together,
(iii) there are always 4 letters between P and S?

EXERCISE 6.4

1If nC8=nC2^n\text{C}_8 = ^n\text{C}_2, find nC2^n\text{C}_2.
2(i)2nC3:nC3=12:1^{2n}\text{C}_3 : ^n\text{C}_3 = 12 : 1
2(ii)2nC3:nC3=11:1^{2n}\text{C}_3 : ^n\text{C}_3 = 11 : 1
3How many chords can be drawn through 21 points on a circle?
4In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
5Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.
6Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
7In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?
8A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
9In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Miscellaneous Exercise on Chapter 6

1How many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER ?
2How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?
3A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of:
(i) exactly 3 girls ? (ii) atleast 3 girls ? (iii) atmost 3 girls ?
4If the different permutations of all the letter of the word EXAMINATION are
5How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated ?
6The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet ?
7In an examination, a question paper consists of 12 questions divided into two parts i.e., Part I and Part II, containing 5 and 7 questions, respectively. A student is required to attempt 8 questions in all, selecting at least 3 from each part. In how many ways can a student select the questions ?
8Determine the number of 5-card combinations out of a deck of 52 cards if each selection of 5 cards has exactly one king.
9It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible ?
10From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can the excursion party be chosen ?
11In how many ways can the letters of the word ASSASSINATION be arranged so that all the S's are together ?

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