There are 35 numbers in the Massachusetts Mass Cash game. In how many ways can a player select five of the numbers?

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Answer 1 · 2017-04-23T18:25:05

Condition where order matters $\to 38955840$ $->38955840$
Condition where order not matter $\to 324632$ $->324632$

Note that the notation ! means 'Factorial'.

An example $4! \to 4 \times 3 \times 2 \times 1$ $4!->4xx3xx2xx1$ where the 1 has no effect.

Case 1
If the order matters then you use the standardised form of $\frac{n!}{(n - r)!}$ $(n!)/((n-r)!)$

Case 2
If order does not matter then we use $\frac{n!}{(n - r)! r!}$ $(n!)/((n-r)!r!)$

The trick is to look for values you can cancel out.
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Suppose we have Case 2

Then $\frac{n!}{(n - r)! r!} \to \frac{35!}{(35 - 5)! 5!}$ $(n!)/((n-r)!r!)" "->" "(35!)/((35-5)!5!)$

Note that $(35 - 5)! \to 30!$ $(35-5)! ->30!$

$(35xx34xx33xx32xx31xxcancel(30!))/ (cancel((30!))xx5xx4xx3xx2xx1)$

$35/5xx34/4xx33/3xx32/2xx31$

$35/5xx34xx33/3xx32/(4xx2)xx31$

$7xx34xx11xx4xx31 = 324632$

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Suppose we have Case 1

Case 2 has part way answered it but we need to cancel out the division by 5! So we multiply by 5!

$324632xx5! = 38955840$