# Mathematics Questions and Answers – Combinations

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Combinations”.

1. Order matters in combination.
a) True
b) False
View Answer

Answer: b
Explanation: Combination means selection and selection does not require ordering. So, order does not matter in combination.

2. nCr = ________________
a) n!
b) $$\frac{n!}{r!}$$
c) $$\frac{n!}{(n-r)!}$$
d) $$\frac{n!}{(n-r)! r!}$$
View Answer

Answer: d
Explanation: Permutation is known as selection. nCr means arranging r objects out of n.
nCr = $$\frac{n!}{(n-r)! r!}$$.

3. nC0 = ________________
a) n!
b) 1
c) $$\frac{1}{(n)!}$$
d) (n-1)!
View Answer

Answer: b
Explanation: We know, nCr = $$\frac{n!}{(n-r)! r!}$$.
nC0 = $$\frac{n!}{(n-0)! 0!} = \frac{n!}{n!(1)}$$ = 1.
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4. nCn = ________________
a) n!
b) 1
c) $$\frac{1}{(n)!}$$
d) (n-1)!
View Answer

Answer: b
Explanation: We know, nCr = $$\frac{n!}{(n-r)! r!}$$.
nCn = $$\frac{n!}{(n-n)! n!} = \frac{n!}{(0)! n!}$$ = 1/1 = 1.

5. nPr = nCr * ______________
a) r!
b) 1/r!
c) n!
d) 1/n!
View Answer

Answer: a
Explanation: We know, nPr = $$\frac{n!}{(n-r)!}$$ and nCr = $$\frac{n!}{(n-r)! r!}$$.
=> nPr / nCr = $$\frac{\frac{n!}{(n-r)!}}{\frac{n!}{(n-r)! r!}}$$ = r!
=> nPr = nCr * r!
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6. Is nCr = nCn-r true?
a) True
b) False
View Answer

Answer: a
Explanation: We know, nCr = $$\frac{n!}{(n-r)! r!}$$.
Replacing r by n-r, we get nCn-r = $$\frac{n!}{(n-(n-r))!(n-r)!} = \frac{n!}{r! (n-r)!}$$ = nCr
=> nCr = nCn-r

7. If nC2 = nC3 then find n.
a) 2
b) 3
c) 5
d) 6
View Answer

Answer: c
Explanation: We know, nCp = nCq
=>either p = q or p + q = n.
Here p=2 and q=3 so, p ≠ q.
Hence p + q = n => n = p + q = 2 + 3 = 5.
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8. If 6C2 = 6CX then find possible values of x.
a) 2
b) 4
c) 2 and 4
d) 3
View Answer

Answer: c
Explanation: We know, nCp = nCq
=>either p = q or p + q = n.
=> either x=2 or x+2=6
=> either x=2 or x=4.

9. Determine n if 2nC3: nC3 = 9:1.
a) 7
b) 14
c) 28
d) 32
View Answer

Answer: b
Explanation: We know, nCr = $$\frac{n!}{(n-r)! r!}$$.
2nC3 / nC3 = $$\frac{\frac{n!}{3! (2n-3)!}}{\frac{n!}{3! (n-3)!}}$$
= $$\frac{2n! (n-3)!}{n! (2n-3)!}$$
9/1 = $$\frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)}$$
9 = $$\frac{2(2n-1)2}{(n-2)}$$
9n-18 = 8n-4
=>n=14.
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10. If 14Cr = 14 and 15Cr = 15. Find the value of 14Cr-1.
a) 1
b) 14
c) 15
d) 3
View Answer

Answer: a
Explanation: We know, nCr + nCr-1 = n+1Cr
Substituting n=14 and r=4, we get 14Cr + 14Cr-1 = 15Cr
=>14Cr-1 = 15Cr14Cr = 15-14 = 1.

11. Out of a group of 5 persons, find the number of ways of selecting 3 persons.
a) 1
b) 5
c) 10
d) 15
View Answer

Answer: c
Explanation: Out of a group of 5 persons, we have to select 3 persons. This can be done in 5C3 ways.
nCr = $$\frac{n!}{(n-r)! r!}$$
5C3 = $$\frac{5!}{(5-3)! 3!} = \frac{5*4*3!}{(2)! 3!} = \frac{20}{2}$$
i.e. 10 ways.

12. In a family, 5 males and 3 females are there. In how many ways we can select a group of 2 males and 2 females from the family?
a) 3
b) 10
c) 30
d) 40
View Answer

Answer: c
Explanation: Out of 5 males, 2 males can be selected in 5C2 ways.
5C2 = $$\frac{5!}{(5-2)! 2!} = \frac{5*4*3!}{(3)! 2!} = \frac{20}{2}$$ = 10 ways.
Out of 3 females, 2 females can be selected in 3C2 ways .
3C2 = $$\frac{3!}{(3-2)! 2!} = \frac{3*2!}{(1)! 2!} = \frac{3}{1}$$ = 3 ways.
So, by the fundamental principle of counting we can select a group of 2 males and 2 females from the family in 10*3 = 30 ways.

Sanfoundry Global Education & Learning Series – Mathematics – Class 11.

To practice all areas of Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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