Combinatorics Basics - IB Mathematics Analysis and Approaches

1. The coefficient of the term $a^3b^7$ in the expansion of $(a+b)^{10}$ is $C$. What is the coefficient of the term $x^7y^3$ in the expansion of $(x+y)^{10}$?
   Answer: $C$ The coefficient of the term $a^k b^{n-k}$ in the expansion of $(a+b)^n$ is given by $\binom{n}{k}$. Or, equivalently, the coefficient of $a^k b^j$ where $k+j=n$ is $\binom{n}{k} = \binom{n}{j}$. For the term $a^3b^7$ in $(a+b)^{10}$, $n=10$, the power of $a$ is 3, and the power of $b$ is 7. The coefficient is $\binom{10}{3}$ or $\binom{10}{7}$. We are given that this value is $C$. So, $C = \binom{10}{3} = \binom{10}{7}$. For the term $x^7y^3$ in the expansion of $(x+y)^{10}$, $n=10$, the power of $x$ is 7, and the power of $y$ is 3. The coefficient is $\binom{10}{7}$ or $\binom{10}{3}$. Since $\binom{10}{7} = \binom{10}{3}$, the coefficient of $x^7y^3$ is also $C$. This question tests the symmetry property of binomial coefficients: $\binom{n}{r} = \binom{n}{n-r}$.
2. Given that $\binom{n}{k-1} = 36$, $\binom{n}{k} = 84$, and $\binom{n}{k+1} = 126$ for some integers $n$ and $k$. Find the value of $\binom{n}{n-k}$.
   Answer: 84 The question asks for the value of $\binom{n}{n-k}$. A key property of binomial coefficients is the symmetry identity: $\binom{n}{r} = \binom{n}{n-r}$. Applying this identity with $r=k$, we get $\binom{n}{k} = \binom{n}{n-k}$. The problem statement gives the value of $\binom{n}{k}$ as 84. Therefore, $\binom{n}{n-k} = 84$. The other values, $\binom{n}{k-1} = 36$ and $\binom{n}{k+1} = 126$, are extra information designed to distract the test-taker.
3. Consider the expansion of $(x^2+1)^m$. The sum of the coefficients of the terms in the expansion is 1024. What is the degree of the resulting polynomial?
   Answer: 20 The sum of the coefficients in any polynomial expansion $P(x)$ can be found by evaluating $P(1)$. For the expansion of $(x^2+1)^m$, the sum of the coefficients is $(1^2+1)^m = (1+1)^m = 2^m$. We are given that this sum is 1024. So, $2^m = 1024$. Since $1024 = 2^{10}$, we can conclude that $m=10$. The polynomial is $(x^2+1)^{10}$. The degree of a polynomial is the highest power of the variable after expansion. The highest power of $x$ will come from the first term in the binomial expansion, which is $\binom{10}{10}(x^2)^{10}(1)^0 = (x^2)^{10} = x^{20}$. Therefore, the degree of the polynomial is 20. A common error is to choose 10, which is the value of $m$, not the degree of the polynomial.
4. The coefficient of $x^5$ in the expansion of $(2x+a)^8$ is 14336. Find the value of $a$.
   Answer: 2 The general term in the expansion of $(2x+a)^8$ is $T_{r+1} = \binom{8}{r}(2x)^r a^{8-r}$. We want the term with $x^5$, so we set $r=5$. The term is $T_6 = \binom{8}{5}(2x)^5 a^{8-5} = \binom{8}{5} 2^5 x^5 a^3$. The coefficient of $x^5$ is $\binom{8}{5} 2^5 a^3$. First, calculate the binomial coefficient: $\binom{8}{5} = \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$. Next, calculate the power of 2: $2^5 = 32$. So the coefficient is $56 \times 32 \times a^3 = 1792 a^3$. We are given that this coefficient is 14336. So, $1792 a^3 = 14336$. Divide to solve for $a^3$: $a^3 = \frac{14336}{1792} = 8$. Taking the cube root, we find $a=2$. Distractor C arises from a calculation error, for instance, using $2^4$ instead of $2^5$, which would lead to $896a^3=14336$, so $a^3=16$, which might be mistaken as $a=4$. Distractor D arises from taking $a=a^3$.
5. In the expansion of $(4+x)^{29}$, the coefficients of $x^k$ and $x^{k+1}$ are equal. Find the value of $k$.
   Answer: 5 The general term in the expansion of $(4+x)^{29}$ is $T_{r+1} = \binom{29}{r} 4^{29-r} x^r$. The coefficient of $x^k$ is obtained when $r=k$, which is $\binom{29}{k} 4^{29-k}$. The coefficient of $x^{k+1}$ is obtained when $r=k+1$, which is $\binom{29}{k+1} 4^{29-(k+1)} = \binom{29}{k+1} 4^{28-k}$. We are given that these coefficients are equal: $\binom{29}{k} 4^{29-k} = \binom{29}{k+1} 4^{28-k}$. Divide both sides by $4^{28-k}$: $\binom{29}{k} \cdot 4 = \binom{29}{k+1}$. Now expand the binomial coefficients using the formula $\binom{n}{r} = \frac{n!}{r!(n-r)!}$: $\frac{29!}{k!(29-k)!} \cdot 4 = \frac{29!}{(k+1)!(29-(k+1))!}$. $\frac{4}{k!(29-k)(28-k)!} = \frac{1}{(k+1)k!(28-k)!}$. Cancel common terms $k!$ and $(28-k)!$: $\frac{4}{29-k} = \frac{1}{k+1}$. Cross-multiply: $4(k+1) = 29-k$. $4k+4 = 29-k$. $5k = 25$. $k=5$. Distractor D (23) comes from incorrectly setting up the ratio as $\frac{1}{29-k} = \frac{4}{k+1}$.
6. In the expansion of $(1+x)^n$, the coefficients of the 4th and 6th terms are equal. What is the value of $n$?
   Answer: 8 In the expansion of $(1+x)^n$, the general term is $T_{r+1} = \binom{n}{r}x^r$. The coefficient is $\binom{n}{r}$. The 4th term corresponds to $r=3$, so its coefficient is $\binom{n}{3}$. The 6th term corresponds to $r=5$, so its coefficient is $\binom{n}{5}$. We are given that these coefficients are equal: $\binom{n}{3} = \binom{n}{5}$. We use the identity $\binom{n}{a} = \binom{n}{b}$ which implies either $a=b$ (not possible here as $3 \neq 5$) or $a+b=n$. Therefore, $n = 3+5 = 8$. Distractor A (n=7) might result from an off-by-one error, comparing $\binom{n}{4}$ and $\binom{n}{6}$ and then making another error. For example, $4+6=10$ and then dividing by something.
7. In the expansion of $(1+x)^n$, the coefficient of $x^2$ is three times the coefficient of $x$. Find the value of $n$.
   Answer: 7 The expansion of $(1+x)^n$ begins $1 + \binom{n}{1}x + \binom{n}{2}x^2 + \dots$. The coefficient of $x$ is $c_1 = \binom{n}{1} = n$. The coefficient of $x^2$ is $c_2 = \binom{n}{2} = \frac{n(n-1)}{2}$. The problem states that $c_2 = 3c_1$. Substituting the expressions for the coefficients, we get: $\frac{n(n-1)}{2} = 3n$. Since $n$ must be at least 2 for an $x^2$ term to exist, $n \neq 0$, so we can divide both sides by $n$: $\frac{n-1}{2} = 3$. Multiply both sides by 2: $n-1 = 6$. Therefore, $n=7$. Distractor A (n=4) would arise if one forgot the denominator 2 in the formula for $\binom{n}{2}$, leading to the incorrect equation $n(n-1)=3n$, which simplifies to $n-1=3$.
8. What is the coefficient of $x$ in the expansion of $\left( x^3 - \frac{2}{x} \right)^7$?
   Answer: -672 The general term in the expansion of $(a+b)^n$ is $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. For $\left( x^3 - \frac{2}{x} \right)^7$, we have $a=x^3$, $b=-\frac{2}{x}$, and $n=7$. The general term is $T_{r+1} = \binom{7}{r} (x^3)^{7-r} \left( -\frac{2}{x} \right)^r$. Let's simplify the expression to find the power of $x$: $T_{r+1} = \binom{7}{r} x^{3(7-r)} (-2)^r x^{-r} = \binom{7}{r} (-2)^r x^{21-3r-r} = \binom{7}{r} (-2)^r x^{21-4r}$. We want the coefficient of $x$, which means the exponent of $x$ should be 1. Set $21-4r = 1$. $20 = 4r$. $r=5$. Now substitute $r=5$ into the coefficient part of the general term: $\binom{7}{5}(-2)^5$. Calculate the parts: $\binom{7}{5} = \binom{7}{2} = \frac{7 \times 6}{2} = 21$ and $(-2)^5 = -32$. The coefficient is $21 \times (-32) = -672$. Distractor D (672) is a common error from forgetting the negative sign on $(-2)^5$.
9. What is the value of the sum $\sum_{r=0}^{10} \binom{10}{r} (-1)^r$?
   Answer: 0 This sum is a direct application of the binomial theorem. The binomial theorem states that $(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$. Let's choose specific values for $a$, $b$, and $n$ to match the given sum. The sum is $\sum_{r=0}^{10} \binom{10}{r} (-1)^r$. We can rewrite this as $\sum_{r=0}^{10} \binom{10}{r} (1)^{10-r} (-1)^r$. This perfectly matches the expansion of $(a+b)^n$ with $n=10$, $a=1$, and $b=-1$. So, the sum is equal to $(1 + (-1))^{10} = (1-1)^{10} = 0^{10} = 0$. In general, for any integer $n \ge 1$, the alternating sum of binomial coefficients $\sum_{r=0}^{n} \binom{n}{r} (-1)^r = 0$. Distractor D (1024) is the value of $\sum_{r=0}^{10} \binom{10}{r} = 2^{10}$.
10. Which of the following expressions is equivalent to $\binom{k}{r} + 2\binom{k}{r+1} + \binom{k}{r+2}$?
    Answer: $\binom{k+2}{r+2}$ This problem uses Pascal's identity, which states that $\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}$. We can rewrite the given expression by splitting the middle term: $\binom{k}{r} + 2\binom{k}{r+1} + \binom{k}{r+2} = \left[ \binom{k}{r} + \binom{k}{r+1} \right] + \left[ \binom{k}{r+1} + \binom{k}{r+2} \right]$. Now, apply Pascal's identity to each pair of terms in the brackets: 1. $\binom{k}{r} + \binom{k}{r+1} = \binom{k+1}{r+1}$ 2. $\binom{k}{r+1} + \binom{k}{r+2} = \binom{k+1}{r+2}$ Substitute these back into the expression: $\binom{k+1}{r+1} + \binom{k+1}{r+2}$. We can apply Pascal's identity one more time to this new expression (with $n=k+1$ and $k=r+1$): $\binom{k+1}{r+1} + \binom{k+1}{r+2} = \binom{(k+1)+1}{(r+1)+1} = \binom{k+2}{r+2}$. Therefore, the original expression is equivalent to $\binom{k+2}{r+2}$.
11. The third term in the expansion of $(1+ax)^n$ is $36x^2$ and the fourth term is $84x^3$. Find the value of $n$.
    Answer: 9 The general term in the expansion of $(1+ax)^n$ is $T_{r+1} = \binom{n}{r}(ax)^r = \binom{n}{r}a^r x^r$. The third term corresponds to $r=2$: $T_3 = \binom{n}{2}a^2x^2$. The coefficient is $\binom{n}{2}a^2 = 36$. The fourth term corresponds to $r=3$: $T_4 = \binom{n}{3}a^3x^3$. The coefficient is $\binom{n}{3}a^3 = 84$. Let's form a ratio of the two coefficients: $\frac{\binom{n}{3}a^3}{\binom{n}{2}a^2} = \frac{84}{36}$. This simplifies to $\frac{\frac{n!}{3!(n-3)!}}{\frac{n!}{2!(n-2)!}} \cdot a = \frac{7}{3}$. $\frac{n!}{3!(n-3)!} \cdot \frac{2!(n-2)!}{n!} \cdot a = \frac{7}{3}$. $\frac{2!(n-2)(n-3)!}{3 \cdot 2!(n-3)!} \cdot a = \frac{7}{3}$. $\frac{n-2}{3} \cdot a = \frac{7}{3}$, which simplifies to $a(n-2) = 7$. From the third term's coefficient: $\frac{n(n-1)}{2}a^2 = 36$. From $a = \frac{7}{n-2}$, we substitute: $\frac{n(n-1)}{2} \left(\frac{7}{n-2}\right)^2 = 36$. $\frac{n(n-1)}{2} \frac{49}{(n-2)^2} = 36$. This looks complicated. Let's test the integer options. If $n=9$, then $a(9-2)=7 \implies 7a=7 \implies a=1$. Check with the third term: $\binom{9}{2}(1)^2 = \frac{9 \times 8}{2} = 36$. This works. Check with the fourth term: $\binom{9}{3}(1)^3 = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$. This also works. Thus, $n=9$.
12. The coefficient of $x^8$ in the expansion of $(x^2+k)^7$ is 280. Find the value of $k$.
    Answer: 2 The general term in the expansion of $(x^2+k)^7$ is $T_{r+1} = \binom{7}{r} (x^2)^{7-r} k^r$. Simplifying the term: $T_{r+1} = \binom{7}{r} x^{14-2r} k^r$. We want the term with $x^8$, so we set the exponent of $x$ to 8. $14-2r = 8$ $6 = 2r$ $r=3$. The coefficient of this term is obtained by substituting $r=3$ into the coefficient part of the general term: $\binom{7}{3}k^3$. Calculate the binomial coefficient: $\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$. The coefficient is $35k^3$. We are given that this coefficient is 280. $35k^3 = 280$. $k^3 = \frac{280}{35} = 8$. Taking the cube root, we find $k=2$.
13. Given that $\binom{n}{2} = 105$, what is the value of $\binom{n-1}{3}$?
    Answer: 364 First, we need to find the value of $n$ from the equation $\binom{n}{2} = 105$. The formula for $\binom{n}{2}$ is $\frac{n(n-1)}{2}$. So, $\frac{n(n-1)}{2} = 105$. $n(n-1) = 210$. We are looking for two consecutive integers whose product is 210. We can estimate: $\sqrt{210}$ is between $\sqrt{196}=14$ and $\sqrt{225}=15$. Let's try $n=15$. Then $n(n-1) = 15(14) = 210$. This is correct, so $n=15$. Now we need to calculate $\binom{n-1}{3}$ with $n=15$. This is $\binom{15-1}{3} = \binom{14}{3}$. $\binom{14}{3} = \frac{14 \times 13 \times 12}{3 \times 2 \times 1}$. $\binom{14}{3} = 14 \times 13 \times 2 = 364$. A common mistake is to calculate $\binom{15}{3}$, which is $455$ (Distractor D).
14. Find the term independent of $x$ in the expansion of $\left( 2x^2 - \frac{1}{x} \right)^9$.
    Answer: 672 The general term of the expansion $(a+b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$. Here, $a = 2x^2$, $b = -\frac{1}{x}$, and $n=9$. The general term is $T_{r+1} = \binom{9}{r} (2x^2)^{9-r} \left(-\frac{1}{x}\right)^r$. Simplifying the term: $T_{r+1} = \binom{9}{r} 2^{9-r} (x^2)^{9-r} (-1)^r (x^{-1})^r = \binom{9}{r} 2^{9-r} (-1)^r x^{18-2r} x^{-r} = \binom{9}{r} 2^{9-r} (-1)^r x^{18-3r}$. The term independent of $x$ is the one where the exponent of $x$ is 0. Set $18-3r = 0$, which gives $3r=18$, so $r=6$. Now substitute $r=6$ back into the expression for the term: Term = $\binom{9}{6} 2^{9-6} (-1)^6 x^{18-3(6)}$ = $\binom{9}{3} 2^3 (1) x^0$. Calculate the value: $\binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$. The term is $84 \times 2^3 = 84 \times 8 = 672$. Distractor A (-5376) could arise from miscalculating the power of x, for example finding r=3, which gives $\binom{9}{3}2^6(-1)^3 = 84 \times 64 \times (-1)$. Distractor B (84) arises from forgetting the $2^{9-r}$ factor. Distractor D could arise from a sign error or miscalculation.

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