Question

# The number of solutions of the equation $$x+y+z=10$$ where $$x,y$$ and $$z$$ are positive integers

A
36
B
55
C
72
D
45

Solution

## The correct option is A $$36$$Given equation is $$x+y+z=10$$where $$x,y$$ and $$z$$ are positive integersFor any pair of positive integers n and k, the number of k-tuples of positive integers whose sum is n is equal to the number of $$(k − 1)$$-element subsets of a set with $$(n − 1)$$ elements.Both of these numbers are given by the binomial coefficient $${\displaystyle \textstyle {n-1 \choose k-1}}.$$$$\therefore$$ Required number of solutions $$={ _{ }^{ (10-1) }{ C } }_{ (3-1) }={ _{ }^{ 9 }{ C } }_{ 2 }=\cfrac { 9\times 8 }{ 2 } =36$$Maths

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