Binomial Coefficient Questions and Answers

This set of Discrete Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Counting – Binomial Coefficient”.

1. Calculate the value of ⁸C₅.
a) 79
b) 43
c) 120
d) 56
View Answer

Answer: d
Explanation: We can use the formula ⁿC_k = \(\frac{n!}{k!(n-k)!}\) 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
View Answer

Answer: c
Explanation: 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
View Answer

Answer: a
Explanation: 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
View Answer

Answer: c
Explanation: the binomial theorem is (x+y)^a = Σ ^aC_i x^a-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
View Answer

Answer: d
Explanation: By using Binomial theorem = ⁿ∑_k=0 (n^k) x^ky^n-k = n⁰x⁰yⁿ + n¹x¹y^n-1 + n²x²y^n-2 + … + nⁿxⁿy⁰, where (n^k) = \(\frac{n!}{k!(n−k)!}\). Now, T_r+1 = ⁿC_ra^n-rb^r, T₉₊₁ = ¹²C₆a^12-6b⁶ = 220 * (3x²)⁶ * (4)⁶ = 220 * 3⁶ * 4⁶. Hence the coefficient is 220 * 3⁶ * 4⁶.

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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) \(\frac{12}{65}\)
View Answer

Answer: a
Explanation: 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
View Answer

Answer: c
Explanation: The coefficient of the 8^th term is ¹¹C₈ = 165. Hence, the 8^th 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
View Answer

Answer: a
Explanation: 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
View Answer

Answer: b
Explanation: 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 51^st 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
View Answer

Answer: d
Explanation: By using the Binomial Theorem, the terms are of the form ⁶C_n * (4x)^6-n * (b/x)ⁿ.
For the term to be independent of x, we need x^6-n(1/x)ⁿ = x⁰ ⇒ x^6-n(x-1)ⁿ = x⁰ ⇒ x^6-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.

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