Question

If $^{30}C_0\ ^{20}C_{10} + ^{31}C_1\ ^{19}C_{10} + ^{32}C_2\ ^{18}C_{10} + .. + ^{40}C_{10}\ ^{10}C_{10} = ^aC_b$, (where a and b are relative co-prime odd numbers) then $\frac{a-b}{2}$ is equal to

Answer 1

5

Answer 2

The given sum is $S = ^{30}C_0\ ^{20}C_{10} + ^{31}C_1\ ^{19}C_{10} + ^{32}C_2\ ^{18}C_{10} + .. + ^{40}C_{10}\ ^{10}C_{10}$. The general term of the sum can be written as $^{30+k}C_k \ ^{20-k}C_{10}$, where $k$ ranges from $0$ to $10$. So, the sum can be written as $S = \sum_{k=0}^{10} \binom{30+k}{k} \binom{20-k}{10}$. Using the identity $\binom{n}{k} = \binom{n}{n-k}$, we have $\binom{20-k}{10} = \binom{20-k}{10-k}$. Thus, $S = \sum_{k=0}^{10} \binom{30+k}{k} \binom{20-k}{10-k}$. This sum matches the form of the combinatorial identity $\sum_{k=0}^{n} \binom{r+k}{k} \binom{s-k}{n-k} = \binom{r+s+1}{n}$. Comparing our sum with the identity, we have $r = 30$, $s = 20$, and $n = 10$. Using the identity, the sum $S = \binom{30+20+1}{10} = \binom{51}{10}$. We are given that $S = ^aC_b$, where $a$ and $b$ are relative co-prime odd numbers. So, $^aC_b = ^{51}C_{10}$. Using the property $\binom{n}{k} = \binom{n}{n-k}$, we have $^{51}C_{10} = ^{51}C_{51-10} = ^{51}C_{41}$. We need to find a pair $(a, b)$ such that $a$ and $b$ are relative co-prime odd numbers. Case 1: $(a, b) = (51, 10)$. Here $a=51$ (odd), but $b=10$ (even). This case is invalid. Case 2: $(a, b) = (51, 41)$. Here $a=51$ (odd) and $b=41$ (odd). The prime factors of 51 are 3 and 17. The prime factor of 41 is 41. Since there are no common prime factors, 51 and 41 are relative co-prime. This case is valid. Therefore, $a=51$ and $b=41$. The question asks for $\frac{a-b}{2}$. $\frac{a-b}{2} = \frac{51-41}{2} = \frac{10}{2} = 5$.