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How many sequences of length 5 can be formed using the digits 0, 1, 2, . . . , 9 with the property that exactly two of the 10 digits appear, e.g., 05550

 Jan 21, 2020
 #1
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If I understand your question !
You want to group the 10 digits from (0 to 9) in groups of 5 numbers in which 2 numbers will be of one kind and the other three of another kind. Example:00111. Is that right?


if so, then let us look at the numbers involved: Let us start with the same example as above: 00111. For the place of the first zero from the left, you have 10 choices. The same goes for the second zero. That is: 10 x 10= 100. The three "1s" can be any of the other 9 numbers, or: 9 x 9 x 9 =729. 


So, you would have: 100 x 729 =72,900. But each group of 5 numbers can have: 5! /3!2! =10 permutations.


Therefore the overall number would be:72,900 x 10 =729,000 such numbers.


Melody: Please check these numbers. Thanks.

 Jan 21, 2020
 #2
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How many sequences of length 5 can be formed using the digits 0, 1, 2, . . . , 9 with the property that exactly two of the 10 digits appear, e.g., 05550

 

Hi Guests,  asker and answerer. laugh

Here is my take. No promises that I am right though.

There are 10C2 = 45  conbinations of 2 numbers

 

You can have them in the ratio 1:4    or   2:3

 

It you have 1 of one of the numbers and 4 of the other then there is 5 ways to do that.

But either can be the single so that is 2 times

so I have   2*5*45 = 450 ways

 

If I have 2 or 1 and 3 of the other then i have 2*5C2ways = 20       

20*45=900 ways

 

900+450 = 1350 combinations that meet that requirement. 

 Jan 22, 2020

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