250+ TOP MCQs on Combinations & Answers | Class 11 Maths

Mathematics Multiple Choice Questions on “Combinations”.

1. Order matters in combination.
a) True
b) False
Answer: b
Clarification: Combination means selection and selection does not require ordering. So, order does not matter in combination.

2. ⁿC_r = ________________
a) n!
b) (frac{n!}{r!})
c) (frac{n!}{(n-r)!})
d) (frac{n!}{(n-r)! r!})
Answer: d
Clarification: Permutation is known as selection. ⁿC_r means arranging r objects out of n.
ⁿC_r = (frac{n!}{(n-r)! r!}).

3. ⁿC₀ = ________________
a) n!
b) 1
c) (frac{1}{(n)!})
d) (n-1)!
Answer: b
Clarification: We know, ⁿC_r = (frac{n!}{(n-r)! r!}).
ⁿC₀ = (frac{n!}{(n-0)! 0!} = frac{n!}{n!(1)}) = 1.

4. ⁿC_n = ________________
a) n!
b) 1
c) (frac{1}{(n)!})
d) (n-1)!
Answer: b
Clarification: We know, ⁿC_r = (frac{n!}{(n-r)! r!}).
ⁿC_n = (frac{n!}{(n-n)! n!} = frac{n!}{(0)! n!}) = 1/1 = 1.

5. ⁿP_r = ⁿC_r * ______________
a) r!
b) 1/r!
c) n!
d) 1/n!
Answer: a
Clarification: We know, ⁿP_r = (frac{n!}{(n-r)!}) and ⁿC_r = (frac{n!}{(n-r)! r!}).
=> ⁿP_r / ⁿC_r = (frac{frac{n!}{(n-r)!}}{frac{n!}{(n-r)! r!}}) = r!
=> ⁿP_r = ⁿC_r * r!

6. Is ⁿC_r = ⁿC_n-r true?
a) True
b) False
Answer: a
Clarification: We know, ⁿC_r = (frac{n!}{(n-r)! r!}).
Replacing r by n-r, we get ⁿC_n-r = (frac{n!}{(n-(n-r))!(n-r)!} = frac{n!}{r! (n-r)!}) = ⁿC_r
=> ⁿC_r = ⁿC_n-r

7. If ⁿC₂ = ⁿC₃ then find n.
a) 2
b) 3
c) 5
d) 6
Answer: c
Clarification: We know, ⁿC_p = ⁿC_q
=>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.

8. If ⁶C₂ = ⁶C_X then find possible values of x.
a) 2
b) 4
c) 2 and 4
d) 3
Answer: c
Clarification: We know, ⁿC_p = ⁿC_q
=>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
Answer: b
Clarification: We know, ⁿC_r = (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.

10. If ¹⁴C_r = 14 and ¹⁵C_r = 15. Find the value of ¹⁴C_r-1.
a) 1
b) 14
c) 15
d) 3
Answer: a
Clarification: We know, ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r
Substituting n=14 and r=4, we get ¹⁴C_r + ¹⁴C_r-1 = ¹⁵C_r
=>¹⁴C_r-1 = ¹⁵C_r – ¹⁴C_r = 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
Answer: c
Clarification: Out of a group of 5 persons, we have to select 3 persons. This can be done in ⁵C₃ ways.
ⁿC_r = (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
Answer: c
Clarification: 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.