combinations: find 1/(2!17!) + 1/(3!16!) + ... + 1/(9!10!)

[malick]

Hello, I have a question.
A competition question a few years back asked to find 1/(2! 17!) + 1/(3! 16!) + ... + 1/(9! 10!).

This was their solution: Put k = 19C2 + 19C3 + ... + 19C9. Then 2k+2(1+19) = 219. So k = 218-20 and sum = (218-20)/19!.

I understant putting k = 19C2 + 19C3 + ... + 19C9, but I don't understand why 2k+2(1+19) = 219.
Can someone explain it to me.

 

[pka]

Well, of course it is not 219 as you have written it but 2<SUP>19</SUP>!
That is due to this fact:
\(\displaystyle \L
2^{19} = \sum\limits_{j = 0}^{19} {\left( \begin{array}{c}
19 \\
j \\
\end{array} \right)} = 1 + 19 + \sum\limits_{j = 2}^9 {\left( \begin{array}{c}
19 \\
j \\
\end{array} \right)} + \sum\limits_{j = 10}^{19} {\left( \begin{array}{c}

19 \\
j \\
\end{array} \right)}
\\

\begin{array}{l}
\left( \begin{array}{c}
19 \\
j \\
\end{array} \right) = \left( \begin{array}{c}
19 \\
19 - j \\
\end{array} \right)\quad \Rightarrow \quad \sum\limits_{j = 10}^{17} {\left( \begin{array}{c}
19 \\
j \\
\end{array} \right)} = \sum\limits_{j = 2}^9 {\left( \begin{array}{c}
19 \\
j \\
\end{array} \right)} \\
\sum\limits_{j = 10}^{19} {\left( \begin{array}{c}
19 \\
j \\
\end{array} \right)} = \sum\limits_{j = 2}^9 {\left( \begin{array}{c}
19 \\
j \\
\end{array} \right)} + 19 + 1 \\
\end{array}\)

 

[soroban]

Re: combinations: find 1/(2!17!) + 1/(3!16!) + ... + 1/(9!10

Hello, malick!

Without looking at their solution, I approached it differently . . .

A competition question asked to find: \(\displaystyle \,S \;= \;\frac{1}{2!17!}\,+\,\frac{1}{3!16!}\,+\,\cdots\,+\,\frac{1}{9!10!}\)

I recognized those fractions as part of the expansion of: \(\displaystyle \,(a\,+\,b)^{19}\)
\(\displaystyle \;\;\)The binomial coefficients are: \(\displaystyle \,C(19,0)\,+\,C(19,1)\,+\,C(19,2)\,+\,\cdots\,+\,C(19,19) \;= \;2^{19}\;\) [1]

Multiply by \(\displaystyle 19!\;\;(19!) S \;= \;\frac{19!}{2!17!}\,+\,\frac{19!}{3!16!}\,+\,\cdots\,+\,\frac{19!}{9!10!}\)

The right side is half of [1], minus the first two terms: \(\displaystyle C(19,19)\,=\,1,\;C(19,1)\,=\,19\)


So we have: \(\displaystyle \,(19!)S\;=\;\frac{1}{2}\left(2^{19})\,-\,1\,-\,19\;=\;2^{18}\,-\,20\)

Therefore: \(\displaystyle \L\,S\;=\;\frac{2^{18}\,-\,20}{19!}\)

 

[malick]

thanks. I believe i get it now.