Problem#20

Relevant wiki: Definition of Binomial Coefficient

$\begin{aligned} \binom{n+k-1}3 & = \binom {n+k}4 & \small \color{#3D99F6} \text{Equating the two binomial coefficients} \\ \frac {(n+k-1)!}{3!(n+k-4)!} & = \frac {(n+k)!}{4!(n+k-4)!} & \small \color{#3D99F6} \text{Note that } \binom MN = \frac {M!}{N!(M-N)!} \\ \frac {\color{#D61F06}\cancel{(n+k-1)!}}{\color{#D61F06}\cancel{3!}\cancel{(n+k-4)!}} & = \frac {\color{#3D99F6}(n+k)\color{#D61F06}\cancel{(n+k-1)!}}{\color{#3D99F6}4 \color{#D61F06} \cancel{(3!)}\cancel{(n+k-4)!}} \\ \implies n+k & = 4 \end{aligned}$

Now, let $\displaystyle P(k) = \binom{n+4}k \binom{k+3}{k+1} = \binom{8-k}k \times \frac {(k+3)!}{2!(k+1)!} = \binom{8-k}k \times \frac {(k+3)(k+2)}2$ . Since $n+k=4$ , $k$ can only take the value from 0 to 4.

$\begin{aligned} P(0) & = \binom 80 \times \frac {3 \times 2}2 = 1\times 3 = 3 \\ P(1) & = \binom 71 \times \frac {4 \times 3}2 = 7 \times 6 = 42 \\ P(2) & = \binom 62 \times \frac {5 \times 4}2 = 15 \times 10 = 150 \\ P(3) & = \binom 53 \times \frac {6 \times 5}2 = 10 \times 15 = 150 \\ P(4) & = \binom 44 \times \frac {7 \times 6}2 = 1 \times 21 = 21 \end{aligned}$

The maximum $P(k) = P(2)=P(3)=\boxed{150}$

From "n choose r" ( click here for background info ), we can rewrite the first two statements like so:

$\frac{(n+k-1)!}{6 \times (n+k-4)!}$ $=$ $\frac{(n+k)!}{24 \times (n+k-4)!}$

Multiplying both sides by $24 \times (n+k-4)!$ gives us:

$4 \times (n+k-1)! = (n+k)!$

From this we can conclude that $n + k = 4$ , since it is the multiplier between two consecutive factorials. We the know that $n = 4 - k$ , allowing us to write the question in terms of $k$ :

$\dbinom{8 - k}{k}\times\dbinom{k+3}{k+1}$

As we know that $n$ and $k$ are whole numbers, $n + k = 4$ , and you can't have the factorial of negative integers, $k = 0,1,2,3,4$

Hence the answer is $\boxed{150}$

$\dbinom{n+k-1}{3}=\dbinom{n+k}{4}\\ \Rightarrow \frac{(n+k-1)!}{3!(n+k-4)!}=\frac{(n+k)!}{4!(n+k-4)!}\\ \Rightarrow n+k=4\\ \therefore n=4-k\\$

$Now \quad \dbinom{n+4}{k}\times\dbinom{k+3}{k+1}\\ =\dbinom{8-k}{k}\times\dbinom{k+3}{k+1}\\ =\frac{(8-k)!}{k!\times(8-2k)!}\times\frac{(k+3)!}{(k+1)!\times(k+3-k-1)!}\\ =\frac{(8-k)!\times(k+3)\times(k+2)}{k!\times(8-2k)!\times2}\\ \\ \\$

Here $k$ must be integer and it's limit is $0\le k\le 4$ . So, $k=0,1,2,3,4$

By inputting $k$ 's value in $\dbinom{n+4}{k}\times\dbinom{k+3}{k+1}=\dbinom{8-k}{k}\times\dbinom{k+3}{k+1}$ ,

we find for $k=2,3$ ; $\dbinom{8-k}{k}\times\dbinom{k+3}{k+1}=\boxed{150}$

and it is the largest.