Discrete Mathematics Multiple Choice Questions on “Counting – Binomial Coefficient”.
1. Calculate the value of 8C5.
a) 79
b) 43
c) 120
d) 56
Answer: d
Clarification: We can use the formula nCk = (frac{n!}{k!(n-k)!}) to calculate the value of 8C5 = (frac{8!}{5!(8-5)!}) = 56.
2. In how many ways can you select 9 cupcakes from a box containing 17 cupcakes?
a) 42769
b) 45398
c) 24310
d) 36214
Answer: c
Clarification: The number of ways to choose 9 cupcakes out of a set of 17 is 17C9 = (frac{17!}{9!(17-9)!}) = 24,310.
3. How many 4-digit numbers can be formed by using 2, 4, 6, 8, 10, 12 without repetition of digits?
a) 15
b) 42
c) 70
d) 127
Answer: a
Clarification: Here making a 4-digit number is equivalent to filling 4 places with 6 numbers. So, the number of ways of filling all the four places is 6C4 = 15. Hence, the total possible 4-digit numbers from the above 6 numbers are 15.
4. What is the coefficient of x9 in the expansion of (x+5)14?
a) 5! * 14C6
b) 14C5
c) 54 * 14C5
d) 34 * 11C5
Answer: c
Clarification: the binomial theorem is (x+y)a = Σ aCi xa-i yi. In order to get the coefficient of x9, we need to have a-i=9. Since a=14, i=5. Thus, the answer is aC5 * y4 = 54 * 14C5.
5. Determine the independent term of x7 in the expansion of (3x2 + 4)12.
a) 220 * 46
b) 230
c) 548* 3!
d) 220 * 36 * 46
Answer: d
Clarification: By using Binomial theorem = n∑k=0 (nk) xkyn-k = n0x0yn + n1x1yn-1 + n2x2yn-2 + … + nnxny0, where (nk) = (frac{n!}{k!(n−k)!}). Now, Tr+1 = nCran-rbr, T9+1 = 12C6a12-6b6 = 220 * (3x2)6 * (4)6 = 220 * 36 * 46. Hence the coefficient is 220 * 36 * 46.
6. In a game, a fair coin is tossed 6 times. Each time the coin comes up tails, A will pay Rs. 15 but if each time heads come up, A will pay nothing. Determine the probability that A will win Rs. 45 by playing the game?
a) (frac{5}{16})
b) (frac{4}{31})
c) (frac{3}{7})
d) (frac{12}{65})
Answer: a
Clarification: By using the binomial distribution, to calculate how likely to win Rs. 45 (or equivalently, the likelihood the coin comes up tails 3 times). The possible outcomes of this game are to win Rs. 45. Therefore, the required probability is (frac{^6C_3}{26} = frac{5}{16}).
7. Find the coefficient of x8 in the expansion of (x+2)11.
a) 640
b) 326
c) 1320
d) 456
Answer: c
Clarification: The coefficient of the 8th term is 11C8 = 165. Hence, the 8th term of the expansion is 165 * 23 * x8 = 1320x8, where the coefficient is 1320.
8. Determine the coefficient of the x5y7 term in the polynomial expansion of (m+n)12.
a) 792
b) 439
c) 382
d) 630
Answer: a
Clarification: Note that, the “x” in the binomial has to be chosen 5 times out of 12. Thus, the coefficient of the term x5y7 must be equal to the number of combinations of 5 objects out of 12: 12C5 = 792.
9. The last digit of the number (((sqrt{51}) + 1)51 – (sqrt{51}) – 1)51 is _______
a) 32
b) 8
c) 51
d) 1
Answer: b
Clarification: Consider the binomial expansion of (m+1)71 and (m-1)71 which gives these two
expressions below respectively: 1) m51 + 51C1m50 + 51C2m49 + 51C3m48 + … + 51C50m1 + 51C51m0
2) m51 – 51C1m50 + 51C2m49 – 51C3m48 + … + 51C50m1 – 51C51m0 .
By taking the difference we have, 2(51C1m50 + 51C3m48 – 51C5m46 + … + 51C50m2 – 51C51m0 ).
In this case, m = (sqrt{51}) and 2(51C1m50 + 51C3m48 – 51C5m46 + … + 51C50m2 – 51C51m0 ).
Consider, module 10 on the powers(for any natural number n): (51)n ≡ (51 mod 10n) ≡ 1 gives 2(51C1 + 51C3 + 51C5 + … + 51C50 + 51C51). Now, by adding the odd terms of the 51st row of the Pascal Triangle 2.((frac{1}{2}) * 251) = 251 = 2(51 mod 4) = 23 = 8.
10. The independent term of x is 80000 in the expansion of (3x+b/x)6, where b is a positive constant. What the value of b?
a) 3.97
b) 6.87
c) 8.3
d) 5.2
Answer: d
Clarification: By using the Binomial Theorem, the terms are of the form 6Cn * (4x)6-n * (b/x)n.
For the term to be independent of x, we need x6-n(1/x)n = x0 ⇒ x6-n(x-1)n = x0 ⇒ x6-nx-n = x0 ⇒ 6 – n = n ⇒ 2n = 6 and n = 3. Thus, we have a constant term of 6C3 * 33 * b3 = 8000
20 * 27 * b3 = 80000
540 * b3 = 80000
b3 = 148.14 ⇒ b= 5.2.