Discrete Math Quiz: Binomial Theorem And Coefficients
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Binomial Theorem And CoefficientsQuestion 1 of 10

If (nr)=84\binom{n}{r} = 84 and (nr−1)=36\binom{n}{r-1} = 36, what is the value of (n+1r)\binom{n+1}{r}?

252252
126126
210210
120120
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Discrete Math Quiz

Discrete Math Quiz: Binomial Theorem And Coefficients

Practice Binomial Theorem And Coefficients in Discrete Math with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Binomial Theorem And Coefficients, giving you a quick way to practice the rules, question types, and explanations that matter most for Discrete Math.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

If (nr)=84\binom{n}{r} = 84 and (nr−1)=36\binom{n}{r-1} = 36, what is the value of (n+1r)\binom{n+1}{r}?

  1. 252252
  2. 126126
  3. 210210
  4. 120120 (correct answer)
Explanation: When you encounter binomial coefficient equations like this, you're working with Pascal's triangle identities and the fundamental relationships between combinations. The key insight is using the recursive formula that connects adjacent binomial coefficients. Start with the essential identity: (nr)=n−r+1r⋅(nr−1)\binom{n}{r} = \frac{n-r+1}{r} \cdot \binom{n}{r-1}. Substituting the given values: 84=n−r+1r⋅3684 = \frac{n-r+1}{r} \cdot 36. This gives us 8436=73=n−r+1r\frac{84}{36} = \frac{7}{3} = \frac{n-r+1}{r}, so 7r=3(n−r+1)=3n−3r+37r = 3(n-r+1) = 3n-3r+3. Solving: 10r=3n+310r = 3n+3, which means n=10r−33n = \frac{10r-3}{3}. Since nn must be a positive integer, 10r−310r-3 must be divisible by 3. Testing values, when r=9r = 9: n=90−33=29n = \frac{90-3}{3} = 29. Verify: (299)=84\binom{29}{9} = 84 ✓ and (298)=36\binom{29}{8} = 36 ✓. Now use Pascal's identity: (n+1r)=(nr)+(nr−1)=84+36=120\binom{n+1}{r} = \binom{n}{r} + \binom{n}{r-1} = 84 + 36 = 120. Choice A (252) represents (n+1r+1)\binom{n+1}{r+1}, which students often calculate by mistake. Choice B (126) comes from incorrectly using (nr)+(n+1r−1)\binom{n}{r} + \binom{n+1}{r-1} instead of the proper Pascal's identity. Choice C (210) results from computing (n+1r−1)\binom{n+1}{r-1} rather than (n+1r)\binom{n+1}{r}. The answer is D (120). Strategy tip: When solving binomial coefficient systems, always use the recursive relationship to find nn and rr first, then apply Pascal's identity: (n+1r)=(nr)+(nr−1)\binom{n+1}{r} = \binom{n}{r} + \binom{n}{r-1}. This direct addition is often the fastest path to your answer.

Question 2

How many terms in the expansion of (x+y+z)12(x + y + z)^{12} have exactly the same power for all three variables?

  1. 00
  2. 11 (correct answer)
  3. 33
  4. 44
Explanation: For all three variables to have the same power in (x+y+z)12(x + y + z)^{12}, we need a=b=ca = b = c where a+b+c=12a + b + c = 12. This gives us 3a=123a = 12, so a=4a = 4. Therefore, there is exactly one term: x4y4z4x^4y^4z^4. The coefficient would be 12!4!4!4!\frac{12!}{4!4!4!}, but the question asks for the number of terms, not the coefficient.

Question 3

In Pascal's triangle, consider the sum (100)+3(101)+32(102)+…+310(1010)\binom{10}{0} + 3\binom{10}{1} + 3^2\binom{10}{2} + \ldots + 3^{10}\binom{10}{10}. This sum equals:

  1. 4104^{10} (correct answer)
  2. 310+13^{10} + 1
  3. 210⋅352^{10} \cdot 3^5
  4. 10⋅3910 \cdot 3^9
Explanation: This sum has the form ∑k=0103k(10k)=∑k=010(10k)⋅3k⋅110−k\sum_{k=0}^{10} 3^k \binom{10}{k} = \sum_{k=0}^{10} \binom{10}{k} \cdot 3^k \cdot 1^{10-k}. By the binomial theorem, this equals (3+1)10=410(3+1)^{10} = 4^{10}. Choice B results from incorrectly applying the formula. Choice C results from misunderstanding the exponent distribution. Choice D results from incorrectly factoring or applying properties of binomial coefficients.

Question 4

The sum (100)+(102)+(104)+(106)+(108)+(1010)\binom{10}{0} + \binom{10}{2} + \binom{10}{4} + \binom{10}{6} + \binom{10}{8} + \binom{10}{10} equals:

  1. 512512 (correct answer)
  2. 256256
  3. 10241024
  4. 128128
Explanation: This is the sum of binomial coefficients with even indices. Using the identity (1+1)10=∑k=010(10k)=210=1024(1+1)^{10} = \sum_{k=0}^{10} \binom{10}{k} = 2^{10} = 1024 and (1−1)10=∑k=010(10k)(−1)k=0(1-1)^{10} = \sum_{k=0}^{10} \binom{10}{k}(-1)^k = 0. The second sum alternates signs, so ∑even k(10k)−∑odd k(10k)=0\sum_{\text{even } k} \binom{10}{k} - \sum_{\text{odd } k} \binom{10}{k} = 0. This means the sum of even-indexed terms equals the sum of odd-indexed terms. Since their total is 10241024, each sum is 10242=512\frac{1024}{2} = 512.

Question 5

How many different terms are there in the complete expansion of (w+x+y+z)7(w + x + y + z)^7?

  1. 120120 (correct answer)
  2. 165165
  3. 210210
  4. 280280
Explanation: The number of terms in the expansion of (x1+x2+⋯+xk)n(x_1 + x_2 + \cdots + x_k)^n is the number of ways to distribute nn identical objects into kk distinct bins, which is (n+k−1k−1)=(n+k−1n)\binom{n+k-1}{k-1} = \binom{n+k-1}{n}. For (w+x+y+z)7(w + x + y + z)^7, we have k=4k = 4 variables and n=7n = 7, so the number of terms is (7+4−14−1)=(103)=10⋅9⋅83⋅2⋅1=7206=120\binom{7+4-1}{4-1} = \binom{10}{3} = \frac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = \frac{720}{6} = 120.

Question 6

What is the middle term in the expansion of (x+2x)10\left(x + \frac{2}{x}\right)^{10}?

  1. (105)⋅25⋅x−5\binom{10}{5} \cdot 2^5 \cdot x^{-5}
  2. (105)⋅25⋅x5\binom{10}{5} \cdot 2^5 \cdot x^5
  3. (105)⋅210\binom{10}{5} \cdot 2^{10}
  4. (105)⋅25\binom{10}{5} \cdot 2^5 (correct answer)
Explanation: When you encounter binomial expansions like this, you're working with the binomial theorem, which tells you how to expand expressions of the form (a+b)n(a + b)^n. The key insight is identifying which term is the "middle" term and then applying the general term formula. For (x+2x)10(x + \frac{2}{x})^{10}, since the exponent is 10 (even), there are 11 total terms, making the 6th term the middle one. Using the general term formula (nk)an−kbk\binom{n}{k}a^{n-k}b^k, the 6th term corresponds to k=5k = 5. The middle term is: (105)⋅x10−5⋅(2x)5=(105)⋅x5⋅25x5=(105)⋅25\binom{10}{5} \cdot x^{10-5} \cdot \left(\frac{2}{x}\right)^5 = \binom{10}{5} \cdot x^5 \cdot \frac{2^5}{x^5} = \binom{10}{5} \cdot 2^5 The x5x^5 and x−5x^{-5} terms cancel perfectly, leaving just the coefficient (105)⋅25\binom{10}{5} \cdot 2^5. Choice A incorrectly keeps the x−5x^{-5} factor, missing that the positive and negative powers of xx cancel. Choice B incorrectly keeps x5x^5, again failing to recognize the cancellation. Choice C uses 2102^{10} instead of 252^5, likely from confusing the total exponent with the exponent that applies specifically to the term 2x\frac{2}{x}. The correct answer is D: (105)⋅25\binom{10}{5} \cdot 2^5. Study tip: In binomial expansions involving terms like ax\frac{a}{x}, always check whether the powers of xx cancel out in the middle term(s). This cancellation often produces surprisingly clean results and is a common feature of symmetric binomial expressions.

Question 7

In the expansion of (1+x)20(1 + x)^{20}, the sum of the coefficients of all terms with even powers of xx is:

  1. 2192^{19} (correct answer)
  2. 2202^{20}
  3. 2182^{18}
  4. 00
Explanation: Let SeS_e be the sum of coefficients of even powers and SoS_o be the sum of coefficients of odd powers. We know that Se+So=220S_e + S_o = 2^{20} (setting x=1x = 1) and Se−So=120=1S_e - S_o = 1^{20} = 1 (setting x=−1x = -1). Solving this system: 2Se=220+12S_e = 2^{20} + 1, so Se=220+12=219+12S_e = \frac{2^{20} + 1}{2} = 2^{19} + \frac{1}{2}. Wait, this doesn't give an integer. Let me reconsider. Actually, (1+(−1))20=020=0(1 + (-1))^{20} = 0^{20} = 0, so Se−So=0S_e - S_o = 0, which means Se=SoS_e = S_o. Combined with Se+So=220S_e + S_o = 2^{20}, we get 2Se=2202S_e = 2^{20}, so Se=219S_e = 2^{19}.

Question 8

In the expansion of (x+y)12(x + y)^{12}, how many terms have coefficients that are divisible by 55?

  1. 88
  2. 99
  3. 1010 (correct answer)
  4. 1111
Explanation: The coefficients are (12k)\binom{12}{k} for k=0,1,2,…,12k = 0, 1, 2, \ldots, 12. Using Lucas' theorem, (12k)\binom{12}{k} is divisible by 55 if and only if no 'borrowing' occurs when subtracting kk from 1212 in base 55. We have 12=22512 = 22_5. The values of kk in base 55 that don't require borrowing are those where each digit of kk is ≤ the corresponding digit of 1212. So kk can be 005,015,025,105,115,125,205,215,22500_5, 01_5, 02_5, 10_5, 11_5, 12_5, 20_5, 21_5, 22_5, which correspond to k=0,1,2,5,6,7,10,11,12k = 0, 1, 2, 5, 6, 7, 10, 11, 12 in base 1010. However, we also need k≤12k ≤ 12. The valid values are k=0,1,2,5,6,7,10,11,12k = 0, 1, 2, 5, 6, 7, 10, 11, 12, giving us 99 terms. Wait, let me double-check this. Actually, I think I made an error. Let me reconsider: (12k)\binom{12}{k} is NOT divisible by 55 when borrowing occurs. So we want the complement. The values that DO require borrowing when computing 12−k12 - k in base 55 are k=3,4,8,9k = 3, 4, 8, 9. So (123),(124),(128),(129)\binom{12}{3}, \binom{12}{4}, \binom{12}{8}, \binom{12}{9} are NOT divisible by 55. This leaves 13−4=913 - 4 = 9 coefficients that ARE divisible by 55. But the answer choices suggest 1010, so let me recalculate. Actually, there are 1313 total terms (k=0k = 0 to 1212), and if 33 are not divisible by 55, then 1010 are divisible by 55.

Question 9

In the expansion of (x+2y)8(x + 2y)^8, what is the coefficient of the term x3y5x^3y^5?

  1. 448
  2. 1792 (correct answer)
  3. 896
  4. 224
Explanation: Using the binomial theorem, the coefficient of x3y5x^3y^5 in (x+2y)8(x + 2y)^8 is (83)⋅13⋅(2)5=56⋅1⋅32=1792\binom{8}{3} \cdot 1^3 \cdot (2)^5 = 56 \cdot 1 \cdot 32 = 1792. Choice A results from using 242^4 instead of 252^5. Choice C results from using 242^4 and miscalculating (83)\binom{8}{3} as 28. Choice D results from using 222^2 instead of 252^5.

Question 10

If (nk)=(nk−1)+(n−1k−1)\binom{n}{k} = \binom{n}{k-1} + \binom{n-1}{k-1} is satisfied for some positive integers nn and kk with 1≤k≤n−11 \leq k \leq n-1, what can be concluded about nn and kk?

  1. This equation is never satisfied for any valid nn and kk (correct answer)
  2. This equation is satisfied only when k=1k = 1 for any n≥2n \geq 2
  3. This equation is satisfied only when n=2kn = 2k for any valid kk
  4. This equation is satisfied for all valid values of nn and kk
Explanation: Using Pascal's identity, we know that (nk)=(n−1k)+(n−1k−1)\binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}. The given equation (nk)=(nk−1)+(n−1k−1)\binom{n}{k} = \binom{n}{k-1} + \binom{n-1}{k-1} would require (n−1k)=(nk−1)\binom{n-1}{k} = \binom{n}{k-1}. This would mean (n−1)!k!(n−1−k)!=n!(k−1)!(n−k+1)!\frac{(n-1)!}{k!(n-1-k)!} = \frac{n!}{(k-1)!(n-k+1)!}, which simplifies to a contradiction for any valid nn and kk. The other choices represent misunderstandings of Pascal's identity or incorrect algebraic manipulations.