Permutations And Combinations (VSAQs)
Maths-2A | 5. Permutations And Combinations – VSAQs:
Welcome to VSAQs in Chapter 5: Permutations And Combinations. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.
VSAQ-1 : If np3=1320 find n
$$1320 = 12 \times 11 \times 10 = 12P3$$
$$n = 12$$
VSAQ-2 : If 12pr=1320 find r
$$1320 = 12 \times 11 \times 10 = 12P3$$
$$r = 3$$
VSAQ-3 : If np4=1680 find n
$$\text{L.H.S} = nP4 = n(n-1)(n-2)(n-3) \ldots (1)$$
$$\text{R.H.S} = 1680 = 168 \times 10$$
$$= (2 \times 84) \times (5 \times 2) = (2 \times 2 \times 42) \times (5 \times 2)$$
$$= (2 \times 2 \times 6 \times 7) \times (5 \times 2) = 8 \times 7 \times 6 \times 5 \ldots (2)$$
Comparing (1) & (2) we get $$n = 8$$
VSAQ-4 : If np7=42. np5 find n
Given that $$nP7 = 42nP5$$
$$\Rightarrow \frac{nP7}{nP5} = \frac{42}{1}$$
$$\Rightarrow \frac{n(n – 1)(n – 2)(n – 3)(n – 4)(n – 5)(n – 6)}{n(n – 1)(n – 2)(n – 3)(n – 4)}$$
$$\Rightarrow (n – 5)(n – 6) = 42$$
$$\Rightarrow (n – 5)(n – 6) = 7 \times 6$$
$$\Rightarrow n – 5 = 7$$
$$\Rightarrow n = 12$$
VSAQ-5 : If (n+1)p5: np5=3:2 then find n
Given that $$(n + 1)P5 : nP5 = 3:2$$
$$\Rightarrow \frac{(n + 1)P5}{nP5} = \frac{3}{2}$$
$$\Rightarrow \frac{(n + 1)n(n – 1)(n – 2)(n – 3)}{n(n – 1)(n – 2)(n – 3)(n – 4)} = \frac{3}{2}$$
$$\Rightarrow 3(n – 4) = 2(n + 1)$$
$$\Rightarrow 3n – 12 = 2n + 2$$
$$\Rightarrow n = 14$$
VSAQ-6 : If (n+1)p7 : np6=2:7 find n
Given that $$(n + 1)P5 : nP6 = 2:7$$
$$\Rightarrow \frac{(n + 1)P5}{nP6} = \frac{2}{7}$$
$$\Rightarrow \frac{(n + 1)n(n – 1)(n – 2)(n – 3)}{n(n – 1)(n – 2)(n – 3)(n – 4)(n – 5)} = \frac{2}{7}$$
$$\Rightarrow 2(n – 4)(n – 5) = 7(n + 1)$$
$$\Rightarrow 2(n^2 – 9n + 20) = 7n + 7$$
$$\Rightarrow 2n^2 – 18n + 40 = 7n + 7$$
$$\Rightarrow 2n^2 – 25n + 33 = 0$$
$$\Rightarrow (n – 11)(2n – 3) = 0$$
$$\Rightarrow n = 11$$
VSAQ-7 : If nC4=210 find n
$$nC4 = 210$$
$$\frac{n(n – 1)(n – 2)(n – 3)}{1 \cdot 2 \cdot 3 \cdot 4} = 10 \times 21$$
$$n(n – 1)(n – 2)(n – 3) = 21 \times 10 \times 1 \times 2 \times 3 \times 4$$
$$= 10 \times 7 \times 3 \times 2 \times 3 \times 4$$
$$= 10 \times 9 \times 8 \times 7$$
$$n = 10$$
VSAQ-8 : If 10.nC2=3.n+1C3 find n
$$10 \cdot nC2 = 3 \cdot {n + 1}C3$$
$$10 \cdot \frac{n(n – 1)}{1 \cdot 2} = 3 \cdot \frac{(n + 1)n(n – 1)}{1 \cdot 2 \cdot 3}$$
$$10 = n + 1$$
$$n = 9$$
VSAQ-9 : If npr=5040 and nCr=210 find n and r
We know that:
$$r! = \frac{nPr}{nCr} = \frac{5040}{210} = 24 = 4!$$
$$nP4 = 5040 = 10 \times 504 = 10 \times 9 \times 56 = 10 \times 9 \times 8 \times 7 = 10P4$$
$$\Rightarrow n = 10$$
$$n = 10, r = 4$$
VSAQ-10 : If nC5 = nC6 then find 13Cn
$$nCr = nCs$$
$$\Rightarrow r + s = n$$ (or) $$r = s$$
$$nC5 = nC6$$
$$\Rightarrow n = 5 + 6 = 11$$
$$13Cn = 13C11 = 13C(13-11)$$
$$= 13C2 = \frac{13 \times 12}{1 \times 2} = 13 \times 6 = 78$$
VSAQ-11 : If nC21= nC27 then find 50Cn
$$nCr = nCs$$
$$\Rightarrow r + s = n$$ (or) $$r = s$$
$$nC21 = nC27$$
$$\Rightarrow n = 21 + 27 = 48$$
$$50Cn = 50C48 = 50C(50-48)$$
$$= 50C2 = \frac{50 \times 49}{1 \times 2} = 25 \times 49 = 1225$$
VSAQ-12 : If 15C2r-1= 15C2r+4 then find r
$$nCr = nCs$$
$$\Rightarrow r + s = n$$ (or) $$r = s$$
$$15C(2r-1) = 15C(2r+4)$$
$$\Rightarrow (2r – 1) + (2r + 4) = 15$$
$$\Rightarrow 4r + 3 = 15$$
$$\Rightarrow 4r = 12$$
$$\Rightarrow r = 3$$
VSAQ-13 : If 12Cr+1= 12C3r-5 find r
$$nCr = nCs$$
$$\Rightarrow r + s = n$$ (or) $$r = s$$
$$12C(r+1) = 12C(3r-5)$$
$$\Rightarrow (r + 1) + (3r – 5) = 12$$
$$\Rightarrow 4r – 4 = 12$$
$$\Rightarrow 4r = 16$$
$$\Rightarrow r = 4$$ (or) $$r + 1 = 3r – 5$$
$$\Rightarrow 2r = 6$$
$$\Rightarrow r = 3$$
VSAQ-14 : Prove that 10C3 + 10C6 = 11C4
L.H.S = $$10 C 3 + 10 C 6 = 10 C 3 + 10 C (10-6)$$
$$10 C 3 + 10 C 4$$
$$11 C 4$$ = R.H.S
$$n C r + n C (r+1) = (n+1) C (r+1)$$
VSAQ-15 : If 9C3 + 9C5 = 10C4 then find r
L.H.S = $$9 C 3 + 9 C 5 = 9 C 3 + 9 C (9-5)$$
$$9 C 3 + 9 C 4$$
$$10 C 4$$
$$n C r + n C (r+1) = (n+1) C (r+1)$$
$$10 C r = 10 C 4$$
$$r = 4$$
VSAQ-16 : Find the value of 10C5 + 2.10C4 + 10C3
$$10 C 5 + 2 \cdot 10 C 4 + 10 C 3$$
$$= 10 C 5 + (10 C 4 + 10 C 4) + 10 C 3$$
$$= (10 C 5 + 10 C 4) + (10 C 4 + 10 C 3)$$
$$= 11 C 5 + 11 C 4$$
$$= 12 C 5 = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792$$
VSAQ-17 : If 12P5 + 5.12P4 = 13Pr find r
We use the formula
$$(n-1) P r + r \cdot (n-1) P (r-1) = n P r$$
$$12 P 5 + 5 \cdot 12 P 4$$
$$= (13-1) P 5 + 5 \cdot (13-1) P (5-1)$$
$$= 13 P 5 = 13 P r$$
$$r = 5$$
VSAQ-18 : Find the number of ways of arranging letters of the word INTERMEDIATE
Given word INTERMEDIATE contains 12 letters
Here, 3 ‘E’s, 2 ‘I’s, 2 ‘T’s are alike
Number of agreements
$$= \frac{n!}{p!q!r!} = \frac{11!}{2!2!2!}$$
VSAQ-19 : Find the number of ways of arranging letters of the word INDEPENDENCE
Given word INDEPENDENCE contains 12 letters
Here, 4 ‘E’s, 3 ‘N’s, 2 ‘D’s are alike
Number of arrangements
$$= \frac{n!}{p!q!r!} = \frac{12!}{4!3!2!}$$
VSAQ-20 : Find the number of ways of arranging letters of the word MATHEMATICS
Given word MATHEMATICS contains 11 letters
Here, 2 ‘M’s, 2 ‘A’s, 2 ‘T’s are alike
Number of arrangements
$$= \frac{n!}{p!q!r!} = \frac{11!}{2!2!2!}$$
VSAQ-21 : Find the number of ways of arranging the letters of the word a4 b3 c5 in its expanded form
The expanded form of a4b3c5 contains 12 letters
Here, 4 ‘a’s, 3 ‘b’s, 5 ‘c’s are alike
Number of arrangements
$$= \frac{n!}{p!q!r!} = \frac{12!}{4!3!5!}$$
VSAQ-22 : Find the number of injections of a set A with 5 elements to a set B with 7 elements
The no of injections from a set containing 5 elements into a set B with 7 elements
$$nPr = 7P5 = 7 \times 6 \times 5 \times 4 \times 3 = 2520$$
VSAQ-23 : Find the number of bijections from set A containing 7 elements onto itself
The number of bijections from a set A containing 7 elements onto itself is
$$7P7 = 7! = 5040$$
VSAQ-24 : Find the number of functions from set A containing 5 elements into a set B containing 4 elements
The no of functions from a set A containing 5 elements into a set B with 4 elements
$$= n^r = 4^5 = 1024$$
VSAQ-25 : Find the number of ways of forming a committee of 5 members from 6 men & 3 ladies
Total number of persons = 6 + 3 = 9
Number of 5 member committee from 9 persons
$$= 9C5 = \frac{9 \times 8 \times 7 \times 6 \times 5}{5 \times 4 \times 3 \times 2 \times 1} = 126$$
VSAQ-26 : Find the number of ways selecting 4 boys and 3 girls from a group of 8 boys and 5 girls
Number of selections of 4 boys from 8 boys = $$8 C 4$$
Number of selections of 3 girls from 5 girls = $$5 C 3$$
Hence, required number of selections = $$8 C 4 \times 5 C 3 = 70 \times 10 = 700$$
VSAQ-27 : Find the number of ways of selecting 3 vowels and 2 consonents from the letters of the word EQUATION
Given word EQUATION contains 5 vowels and 3 consonents
Hence, the number of selections of 3 vowels and 2 consonents
$$= 5 C 3 \times 3 C 2 = 10 \times 3 = 30$$
VSAQ-28 : Find the number of different chains that can be prepared using 6 different coloured beads
Number of circular permutations from n things = $$\frac{1}{2} (n – 1)!$$
Hence the number of chains
$$= \frac{1}{2} (6 – 1)! = \frac{1}{2} (5!) = \frac{1}{2} (120) = 60$$
VSAQ-29 : Find the number of different chains that can be prepared using 7 different coloured beads
Number of circular permutations from n things = $$\frac{1}{2} (n – 1)!$$
Hence the number of chains
$$= \frac{1}{2} (7 – 1)! = \frac{1}{2} (6!) = \frac{1}{2} (720) = 360$$
VSAQ-30 : Find the number of diagonals of a polygon with 12 sides
Number of sides $$n = 12$$
Number of diagonals in ‘n’ – gon
$$= \frac{n(n – 3)}{2}$$
$$= \frac{12 \times 9}{2}$$
$$= 54$$
VSAQ-31 : Find the number of positive divisors of 1080
$$1080 = 108 \times 10 = (2 \times 54) \times (2 \times 5)$$
$$= (2 \times 2 \times 27) \times (2 \times 5)$$
$$= (2 \times 2 \times 3 \times 9) \times (2 \times 5)$$
$$= 2^3 \times 3^3 \times 5^1$$
The number of positive divisors:
$$= (3 + 1)(3 + 1)(1 + 1)$$
$$= 4 \times 4 \times 2$$
$$= 32$$
VSAQ-32 : A man has 4 sons and there are 5 schools within his reach. In how many ways can he admit his sons in the schools so that no two of them will be in the same school
The required number of admissions
$$n P r = 5 P 4$$
$$= 5 \times 4 \times 3 \times 2$$
$$= 120$$
VSAQ-33 : In a class there are 30 students. If each student plays a chess game with each of the other student, then find the total number of chess games played by them
Total number of students = 30
One chess game requires two students to play
Hence order of players (A with B and B with A are same) has no importance
Total number of chess games
$$n C 2 = 30 C 2 = \frac{30 \times 29}{2} = 15 \times 29 = 435$$