Discrete Mathematics Questions and Answers – Counting – Binomial Coefficient

«
»

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 nCk = \(\frac{n!}{k!(n-k)!}\) to calculate the value of 8C5 = \(\frac{8!}{5!(8-5)!}\) = 56.
advertisement

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 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
View Answer

Answer: d
Explanation: By using Binomial theorem = nk=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.
advertisement

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 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
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) m5151C1m50 + 51C2m4951C3m48 + … + 51C50m151C51m0 .
By taking the difference we have, 2(51C1m50 + 51C3m4851C5m46 + … + 51C50m251C51m0 ).
In this case, m = \(\sqrt{51}\) and 2(51C1m50 + 51C3m4851C5m46 + … + 51C50m251C51m0 ).
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.
advertisement

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 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.

Sanfoundry Global Education & Learning Series – Discrete Mathematics.

To practice all areas of Discrete Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

advertisement
advertisement
advertisement
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