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I had another problem, where I got the expression

$$p_k=\frac{\frac{5!}{(5-(k-1))!}}{\frac{8!}{(8-(k-1))!}}\cdot\frac{3}{8-(k-1)}.$$

Using software, this should simplify to

$$p_k=\frac{(k-7)(k-8)}{112}.$$

Question 1: Since I'm not allowed to use software on exams, I wonder how one can simplify this by hand?

Question 2: After having simplified $p_k$, how do I compute

$$\sum_{k=1}^6 p_k$$

by hand?

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    $\begingroup$ Esentially, it amounts to use this factorial property: $n! = n\left(n - 1\right)\cdots\left(n - k + 1\right)\left(n - k\right)!$ $\endgroup$ Apr 26, 2018 at 21:46

1 Answer 1

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$$p_k=\frac {5!}{8!}\frac {(9-k)!}{(6-k)!}\frac {3}{9-k}$$ $$=\frac {1}{6.7.8}\frac {3(9-k)(8-k)(7-k)}{9-k}=\frac {(8-k)(7-k)}{2.7.8}. $$

Once you have the expression of $p_k $,
observe that

$$(8-k)(7-k)=k^2-15k+56.$$ thus

$$112\sum_{k=1}^6p_k=$$ $$ (\sum_{k=1}^6k^2-15\sum_{k=1}^6 k+56\times 6)=$$

$$(\frac {6.7.13}{6}-15\frac {6\times 7}{2}+336)=$$ $$(91-315+336)=112$$

Your sum is $$\sum_{k=1}^6p_k=1.$$

We used the well known sums $$\sum_{k=1}^nk=\frac {n (n+1)}{2} $$ and $$\sum_{k=1}^nk^2=\frac {n (n+2)(2n+1)}{6} $$

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