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September 17th, 2018, 09:12 AM  #1 
Member Joined: Sep 2014 From: Sweden Posts: 94 Thanks: 0  How many different 5digit number combinations can be created?
How many different 5digit 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? 
September 17th, 2018, 10:07 AM  #2 
Senior Member Joined: Feb 2010 Posts: 697 Thanks: 135 
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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