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.
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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.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter