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September 17th, 2018, 09:12 AM   #1
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Question How many different 5-digit number combinations can be created?

How many different 5-digit number combinations can be created using 5, 7ths and 3, 2s?

77777 | 5
222 | 3

So my first thought was to use this formula:

$\displaystyle n_{1} * n_{2} * ... * n_{n}$

So that I get:
$\displaystyle 5! = 5 * 4 * 3 * 2 * 1$
but this is probably just calculating how
many ways I can put all 7s. And I only
have 7s so it do not tell me much.

Then I would do the same thing for the 3, 2s.
And that can't work. I thought if I added
the two results that might give me the
answer, but I think it doesn't since I only
find out how many combinations it is possible
to put 5, 7s in and 3, 2s in.

So I thought the problem might want me to
use this formula:

$\displaystyle \frac{n!}{(n - k)!}$

So how do you actually begin to solve this?
DecoratorFawn82 is offline  
September 17th, 2018, 10:07 AM   #2
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You have five 7's and three 2's and wish to construct a five digit number.

One way is to directly count cases
five 7's and zero 2's for a total of $\displaystyle \dfrac{5!}{5!0!}=1$ way
four 7's and one 2 for a total of $\displaystyle \dfrac{5!}{4!1!}=5$ ways
three 7's and two 2's for a total of $\displaystyle \dfrac{5!}{3!2!}=10$ ways
two 7's and three 2's for a total of $\displaystyle \dfrac{5!}{2!3!}=10$ ways

Adding gives $\displaystyle 1+5+10+10=26$ ways.

The only other way I know is to use an exponential generating function. Go look it up.
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