In how many ways can five positive integers be added such that their sums ten?

Question

Question

Answer 1

Answer #1

here let 5 integers are x1,x2,x3,x4 and x5

therefore x1+x2+x3+x4+x5 =10

number of solution of above equation = $1$ = $5 3$ = $inom{9}{4}$ =9!/(4!*5!)=126

Answer 2

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