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

Guest Jan 21, 2020

#1**+1 **

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.**

Guest Jan 21, 2020

#2**+1 **

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.

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.

Melody Jan 22, 2020