Question

$^{28}C_4 = ^9C_b$ Find max value of $a+b$.

• A. $32$
• B. $52$
• C. $13$
• D. $14$

Answer 1

Answer: (B)

52

Answer 2

The given equation is $^{28}C_4 = ^9C_b$. We are asked to find the maximum value of $a+b$.

First, let's calculate the value of $^{28}C_4$: $^{28}C_4 = \frac{28!}{4!(28-4)!} = \frac{28 \times 27 \times 26 \times 25}{4 \times 3 \times 2 \times 1} = 7 \times 9 \times 13 \times 25 = 20475$

Now, the equation becomes: $20475 = ^9C_b$

The maximum value $^9C_b$ can take is $^9C_4$ or $^9C_5$, which is 126. Since $20475 > 126$, there is no integer solution for $b$ in the given equation.

This indicates a likely typo in the question. A common pattern in such problems is that the bases of the combinations are the same. Let's assume the question was intended to be: $^{28}C_4 = ^{28}C_b$ In this context, we can infer that $a=28$.

From the property of combinations, if $^nC_r = ^nC_s$, then either $r=s$ or $r+s=n$. Applying this to $^{28}C_4 = ^{28}C_b$:

1. $b = 4$
2. $b = 28 - 4 = 24$

We need to find the maximum value of $a+b$. Case 1: $a=28$ and $b=4$. Then $a+b = 28+4 = 32$. Case 2: $a=28$ and $b=24$. Then $a+b = 28+24 = 52$.

The maximum value of $a+b$ is $52$.