Let $$\binom{n}{k}$$ denote $$^nC_k$$ and $$\left[\frac{n}{k}\right] = \begin{cases} \binom{n}{k}, & \text{if } 0 \leq k \leq n \\ 0, & \text{otherwise} \end{cases}$$. If $$A_k = \sum_{i=0}^{9} \binom{9}{i} \left[\binom{12}{12-k+i}\right] + \sum_{i=0}^{8} \binom{8}{i} \left[\binom{13}{13-k+i}\right]$$ and $$A_4 - A_3 = 190p$$, then $$p$$ is equal to _________

We have been given the notation

$$\left[\frac{n}{k}\right]=\begin{cases} \displaystyle\binom{n}{k}, & 0\le k\le n\\[4pt] 0, & \text{otherwise} \end{cases}$$

and the expression

$$A_k=\sum_{i=0}^{9}\binom{9}{i}\left[\frac{12}{\,12-k+i\,}\right]+\sum_{i=0}^{8}\binom{8}{i}\left[\frac{13}{\,13-k+i\,}\right].$$

Only those terms survive in which the lower argument of the bracket lies between $$0$$ and the upper argument inclusive. Thus, for the first summation the condition is

$$0\le 12-k+i\le 12\quad\Longrightarrow\quad k-12\le i\le k,$$

and for the second summation the condition is

$$0\le 13-k+i\le 13\quad\Longrightarrow\quad k-13\le i\le k.$$

Since $$i$$ is non-negative in both sums, the effective range is always $$0\le i\le k$$ (because $$k\le 13$$ in our problem). We shall now evaluate $$A_3$$ and $$A_4$$ separately.

1. Evaluation of $$A_3$$

When $$k=3$$ the admissible indices are $$i=0,1,2,3$$.

First sum — put $$k=3$$:

$$\left[\frac{12}{\,12-3+i\,}\right]=\binom{12}{\,9+i\,},$$ so $$\sum_{i=0}^{3}\binom{9}{i}\binom{12}{\,9+i\,}$$

Term-wise calculation:

$$\begin{aligned} i=0:&\; \binom{9}{0}\binom{12}{9}=1\cdot220=220\\ i=1:&\; \binom{9}{1}\binom{12}{10}=9\cdot66=594\\ i=2:&\; \binom{9}{2}\binom{12}{11}=36\cdot12=432\\ i=3:&\; \binom{9}{3}\binom{12}{12}=84\cdot 1=84 \end{aligned}$$

Adding,

$$220+594+432+84=1330.$$

Second sum — put $$k=3$$:

$$\left[\frac{13}{\,13-3+i\,}\right]=\binom{13}{\,10+i\,},$$ so $$\sum_{i=0}^{3}\binom{8}{i}\binom{13}{\,10+i\,}$$

Term-wise calculation:

$$\begin{aligned} i=0:&\; \binom{8}{0}\binom{13}{10}=1\cdot286=286\\ i=1:&\; \binom{8}{1}\binom{13}{11}=8\cdot78=624\\ i=2:&\; \binom{8}{2}\binom{13}{12}=28\cdot13=364\\ i=3:&\; \binom{8}{3}\binom{13}{13}=56\cdot 1=56 \end{aligned}$$

Adding,

$$286+624+364+56=1330.$$

Therefore

$$A_3=1330+1330=2660.$$

2. Evaluation of $$A_4$$

When $$k=4$$ the admissible indices are $$i=0,1,2,3,4$$.

First sum — put $$k=4$$:

$$\left[\frac{12}{\,12-4+i\,}\right]=\binom{12}{\,8+i\,},$$ so $$\sum_{i=0}^{4}\binom{9}{i}\binom{12}{\,8+i\,}$$

Term-wise calculation:

$$\begin{aligned} i=0:&\; \binom{9}{0}\binom{12}{8}=1\cdot495=495\\ i=1:&\; \binom{9}{1}\binom{12}{9}=9\cdot220=1980\\ i=2:&\; \binom{9}{2}\binom{12}{10}=36\cdot66=2376\\ i=3:&\; \binom{9}{3}\binom{12}{11}=84\cdot12=1008\\ i=4:&\; \binom{9}{4}\binom{12}{12}=126\cdot 1=126 \end{aligned}$$

Adding,

$$495+1980+2376+1008+126=5985.$$

Second sum — put $$k=4$$:

$$\left[\frac{13}{\,13-4+i\,}\right]=\binom{13}{\,9+i\,},$$ so $$\sum_{i=0}^{4}\binom{8}{i}\binom{13}{\,9+i\,}$$

Term-wise calculation:

$$\begin{aligned} i=0:&\; \binom{8}{0}\binom{13}{9}=1\cdot715=715\\ i=1:&\; \binom{8}{1}\binom{13}{10}=8\cdot286=2288\\ i=2:&\; \binom{8}{2}\binom{13}{11}=28\cdot78=2184\\ i=3:&\; \binom{8}{3}\binom{13}{12}=56\cdot13=728\\ i=4:&\; \binom{8}{4}\binom{13}{13}=70\cdot 1=70 \end{aligned}$$

Adding,

$$715+2288+2184+728+70=5985.$$

Therefore

$$A_4=5985+5985=11970.$$

3. Required difference

We now compute

$$A_4-A_3=11970-2660=9310.$$

The question states that

$$A_4-A_3=190\,p,$$ so $$p=\frac{9310}{190}=49.$$

Hence, the correct answer is Option 49.