Combinations

How many solutions are there to x_1+x_2+x_3+x_4=25 where x_n are all non-negative and x_1 is at most 5?
Not sure where to start

could u replace x_4 as x_4'+10 to get x_1+x_2+x_3+x_4'=20

Then get... 23C3 from combinations formula for repetition is not allowed and order is irrelevant

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Originally Posted by qwerty10

How many solutions are there to x_1+x_2+x_3+x_4=25 where x_n are all non-negative and x_1 is at most 5?

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There are $\displaystyle \dbinom{19+4-1}{19}$ solutions in which $\displaystyle x_1\ge6$.
So now what do we do?

would u get: 25+4-1C4-1 - 19+4-1C4-1?

Quote:

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Originally Posted by qwerty10

would u get: 25+4-1C4-1 - 19+4-1C4-1?

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Yes you would. Note I changed the 5 to a 4 in my OP.