Question:

Step 1: Simplify the sum $S$. The given sum is $S = \displaystyle\sum_{r=0}^{n}\binom{n}{r}\binom{n}{n-r+1}$. 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: $$ S = \displaystyle\sum_{r=0}^{n}\binom{n}{r}\binom{n}{r-1} $$ This sum represents the coefficient of $x^{n+1}$ in the expansion of $(1+x)^n (1+x)^n = (1+x)^{2n}$. By Vandermonde's Identity, $\displaystyle\sum_{k=0}^{m}\binom{A}{k}\binom{B}{m-k} = \binom{A+B}{m}$. In our sum, let $A=n$, $B=n$, and the sum of the lower indices is $r + (n-r+1) = n+1$. Thus, $S = \binom{n+n}{n+1} = \binom{2n}{n+1}$. Step 2: Substitute $n=10$. For $n=10$, the value of $S$ is: $$ S = \binom{2 \times 10}{10+1} = \binom{20}{11} $$ Step 3: Determine the number of trailing zeros in $S$. The number of trailing zeros in an integer is determined by the number of factors of 10 in its prime factorization, which is the minimum of the number of factors of 2 and 5. We use Legendre's formula, $v_p(k!) = \sum_{i=1}^{\infty} \lfloor k/p^i \rfloor$. For $S = \binom{20}{11} = \frac{20!}{11!9!}$: Number of factors of 5 in $S$: $v_5(20!) = \lfloor 20/5 \rfloor + \lfloor 20/25 \rfloor = 4 + 0 = 4$. $v_5(11!) = \lfloor 11/5 \rfloor = 2$. $v_5(9!) = \lfloor 9/5 \rfloor = 1$. The number of factors of 5 in $S$ is $v_5(S) = v_5(20!) - v_5(11!) - v_5(9!) = 4 - (2+1) = 4 - 3 = 1$. Number of factors of 2 in $S$: $v_2(20!) = \lfloor 20/2 \rfloor + \lfloor 20/4 \rfloor + \lfloor 20/8 \rfloor + \lfloor 20/16 \rfloor = 10 + 5 + 2 + 1 = 18$. $v_2(11!) = \lfloor 11/2 \rfloor + \lfloor 11/4 \rfloor + \lfloor 11/8 \rfloor = 5 + 2 + 1 = 8$. $v_2(9!) = \lfloor 9/2 \rfloor + \lfloor 9/4 \rfloor + \lfloor 9/8 \rfloor = 4 + 2 + 1 = 7$. The number of factors of 2 in $S$ is $v_2(S) = v_2(20!) - v_2(11!) - v_2(9!) = 18 - (8+7) = 18 - 15 = 3$. The number of trailing zeros in $S$ is $\min(v_5(S), v_2(S)) = \min(1, 3) = 1$. Therefore, $S$ ends in exactly one zero. Step 4: Determine the number of zeros at the end of $11S$. If an integer $S$ ends in $k$ zeros, it can be written as $S = A \times 10^k$, where $A$ is an integer not divisible by 10. Then $11S = 11 \times (A \times 10^k) = (11A) \times 10^k$. Since 11 is not divisible by 2 or 5, and $A$ is not divisible by 2 or 5, their product $11A$ is also not divisible by 2 or 5. Thus, $11S$ will also end in $k$ zeros. Since $S$ ends in 1 zero, $11S$ also ends in 1 zero. The number of zeros at the end of the value of $(11S)$ is 1.