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

1. Calculate the value of ^{8}C_{5}.

a) 79

b) 43

c) 120

d) 56

View Answer

Explanation: We can use the formula

^{n}C

_{k}= \(\frac{n!}{k!(n-k)!}\) to calculate the value of

^{8}C

_{5}= \(\frac{8!}{5!(8-5)!}\) = 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

Explanation: The number of ways to choose 9 cupcakes out of a set of 17 is

^{17}C

_{9}= \(\frac{17!}{9!(17-9)!}\) = 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

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

^{6}C

_{4}= 15. Hence, the total possible 4-digit numbers from the above 6 numbers are 15.

4. What is the coefficient of x^{9} in the expansion of (x+5)^{14}?

a) 5! * ^{14}C_{6}

b) ^{14}C_{5}

c) 54 * ^{14}C_{5}

d) 34 * ^{11}C_{5}

View Answer

Explanation: the binomial theorem is (x+y)

^{a}= Σ

^{a}C

_{i}x

^{a-i}y

^{i}. In order to get the coefficient of x

^{9}, we need to have a-i=9. Since a=14, i=5. Thus, the answer is

^{a}C

_{5}* y

^{4}= 5

^{4}*

^{14}C

_{5}.

5. Determine the independent term of x^{7} in the expansion of (3x^{2} + 4)^{12}.

a) 220 * 4^{6}

b) 230

c) 548* 3!

d) 220 * 3^{6} * 4^{6}

View Answer

Explanation: By using Binomial theorem =

^{n}∑

_{k=0}(n

^{k}) x

^{k}y

^{n-k}= n

^{0}x

^{0}y

^{n}+ n

^{1}x

^{1}y

^{n-1}+ n

^{2}x

^{2}y

^{n-2}+ … + n

^{n}x

^{n}y

^{0}, where (n

^{k}) = \(\frac{n!}{k!(n−k)!}\). Now, T

_{r+1}=

^{n}C

_{r}a

^{n-r}b

^{r}, T

_{9+1}=

^{12}C

_{6}a

^{12-6}b

^{6}= 220 * (3x

^{2})

^{6}* (4)

^{6}= 220 * 3

^{6}* 4

^{6}. Hence the coefficient is 220 * 3

^{6}* 4

^{6}.

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

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 x^{8} in the expansion of (x+2)^{11}.

a) 640

b) 326

c) 1320

d) 456

View Answer

Explanation: The coefficient of the 8

^{th}term is

^{11}C

_{8}= 165. Hence, the 8

^{th}term of the expansion is 165 * 2

^{3}* x

^{8}= 1320x

^{8}, where the coefficient is 1320.

8. Determine the coefficient of the x^{5}y^{7} term in the polynomial expansion of (m+n)^{12}.

a) 792

b) 439

c) 382

d) 630

View Answer

Explanation: Note that, the “x” in the binomial has to be chosen 5 times out of 12. Thus, the coefficient of the term x

^{5}y

^{7}must be equal to the number of combinations of 5 objects out of 12:

^{12}C

_{5}= 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

Explanation: Consider the binomial expansion of (m+1)

^{71}and (m-1)

^{71}which gives these two

expressions below respectively: 1) m

^{51}+

^{51}C

_{1}m

^{50}+

^{51}C

_{2}m

^{49}+

^{51}C

_{3}m

^{48}+ … +

^{51}C

_{50}m

^{1}+

^{51}C

_{51}m

^{0 }

2) m

^{51}–

^{51}C

_{1}m

^{50}+

^{51}C

_{2}m

^{49}–

^{51}C

_{3}m

^{48}+ … +

^{51}C

_{50}m

^{1}–

^{51}C

_{51}m

^{0 }.

By taking the difference we have, 2(

^{51}C

_{1}m

^{50}+

^{51}C

_{3}m

^{48}–

^{51}C

_{5}m

^{46}+ … +

^{51}C

_{50}m

^{2}–

^{51}C

_{51}m

^{0 }).

In this case, m = \(\sqrt{51}\) and 2(

^{51}C

_{1}m

^{50}+

^{51}C

_{3}m

^{48}–

^{51}C

_{5}m

^{46}+ … +

^{51}C

_{50}m

^{2}–

^{51}C

_{51}m

^{0 }).

Consider, module 10 on the powers(for any natural number n): (51)

^{n}≡ (51 mod 10

^{n}) ≡ 1 gives 2(

^{51}C

_{1}+

^{51}C

_{3}+

^{51}C

_{5}+ … +

^{51}C

_{50}+

^{51}C

_{51}). Now, by adding the odd terms of the 51

^{st}row of the Pascal Triangle 2.(\(\frac{1}{2}\) * 2

^{51}) = 2

^{51}= 2

^{(51 mod 4)}= 2

^{3}= 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

Explanation: By using the Binomial Theorem, the terms are of the form

^{6}C

_{n}* (4x)

^{6-n}* (b/x)

^{n}.

For the term to be independent of x, we need x

^{6-n}(1/x)

^{n}= x

^{0}⇒ x

^{6-n}(x-1)

^{n}= x

^{0}⇒ x

^{6-n}x

^{-n}= x

^{0}⇒ 6 – n = n ⇒ 2n = 6 and n = 3. Thus, we have a constant term of

^{6}C

_{3}* 3

^{3}* b

^{3}= 8000

20 * 27 * b

^{3}= 80000

540 * b

^{3}= 80000

b

^{3}= 148.14 ⇒ b= 5.2.

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