Question:

Step 1: Simplify the sum $S$. The sum is given by $S = \displaystyle\sum_{r=0}^{n}\binom{n}{r}\binom{n}{n-r+1}$. Using the identity $\binom{n}{k} = \binom{n}{n-k}$, we can rewrite the second binomial coefficient: $$ \binom{n}{n-r+1} = \binom{n}{n-(n-r+1)} = \binom{n}{r-1} $$ Substituting this into the sum, we get: $$ S = \displaystyle\sum_{r=0}^{n}\binom{n}{r}\binom{n}{r-1} $$ Since $\binom{n}{r-1}=0$ for $r-1 < 0$ (i.e., $r<1$), the sum effectively starts from $r=1$: $$ S = \displaystyle\sum_{r=1}^{n}\binom{n}{r}\binom{n}{r-1} $$ This sum is a form of Vandermonde's Identity. It represents the coefficient of $x^{n+1}$ in the product of two expansions of $(1+x)^n$. Consider the product $(1+x)^n (1+x)^n = (1+x)^{2n}$. The coefficient of $x^{n+1}$ in $(1+x)^{2n}$ is $\binom{2n}{n+1}$. By Vandermonde's Identity, $\sum_{k=0}^{m} \binom{A}{k}\binom{B}{m-k} = \binom{A+B}{m}$. Setting $A=n$, $B=n$, and $m=n+1$, the sum is $\sum_{k=0}^{n+1} \binom{n}{k}\binom{n}{n+1-k}$. Since $\binom{n}{k}=0$ for $k>n$, this sum is equivalent to $\sum_{k=0}^{n} \binom{n}{k}\binom{n}{n+1-k}$. Thus, $S = \binom{2n}{n+1}$. Step 2: Calculate $S$ for $n=10$. Given $n=10$, we substitute this value into the expression for $S$: $$ S = \binom{2 \times 10}{10+1} = \binom{20}{11} $$ Step 3: Determine the number of factors of 5 and 2 in $S$. The number of trailing zeros in an integer is determined by the minimum count of its prime factors 2 and 5. We use Legendre's formula, $N_p(k!) = \sum_{i=1}^{\infty} \lfloor k/p^i \rfloor$, to find the number of prime factors $p$ in $k!$. For $S = \binom{20}{11} = \frac{20!}{11!9!}$: Number of factors of 5 in $S$: $N_5(20!) = \lfloor 20/5 \rfloor = 4$. $N_5(11!) = \lfloor 11/5 \rfloor = 2$. $N_5(9!) = \lfloor 9/5 \rfloor = 1$. The number of factors of 5 in $S$ is $N_5(S) = N_5(20!) - (N_5(11!) + N_5(9!)) = 4 - (2+1) = 1$. Number of factors of 2 in $S$: $N_2(20!) = \lfloor 20/2 \rfloor + \lfloor 20/4 \rfloor + \lfloor 20/8 \rfloor + \lfloor 20/16 \rfloor = 10+5+2+1 = 18$. $N_2(11!) = \lfloor 11/2 \rfloor + \lfloor 11/4 \rfloor + \lfloor 11/8 \rfloor = 5+2+1 = 8$. $N_2(9!) = \lfloor 9/2 \rfloor + \lfloor 9/4 \rfloor + \lfloor 9/8 \rfloor = 4+2+1 = 7$. The number of factors of 2 in $S$ is $N_2(S) = N_2(20!) - (N_2(11!) + N_2(9!)) = 18 - (8+7) = 3$. Step 4: Determine the number of factors of 5 and 2 in $11S$. The value $11S$ is $11 \times S$. The prime factor 11 does not contribute any factors of 2 or 5. Therefore, the number of factors of 5 in $11S$ is $N_5(11S) = N_5(S) = 1$. The number of factors of 2 in $11S$ is $N_2(11S) = N_2(S) = 3$. Step 5: Conclude the number of zeros at the end of $11S$. The number of zeros at the end of $11S$ is given by the minimum of the counts of its prime factors 2 and 5: $$ \text{Number of zeros} = \min(N_2(11S), N_5(11S)) = \min(3, 1) = 1 $$ Thus, the value of $(11S)$ has 1 zero at its end.