Question

Suppose you are shopping for fruit at the grocery store. There are apples, bananas and oranges. In how many ways can you buy 10 items of fruit? For example, you can buy 5 bananas, 3 oranges, and 2 apples; you could also buy 10 bananas and 0 apples and oranges. In how many ways can you buy 10 items of fruit, where you have to have at least one banana, one orange, and one apple?

Answer 1

1)here let x1,x2 and x3 are number of apples,bananas and oranges

therefore x1+x2+x3 =10

number of non-negative solution of above =$\binom{10+3-1}{3-1}$ =$\binom{12}{2}$ =12!/(2!*10!)=66

2)

here as at least one item to be distributed in each category; therefore

x1+x2+x3 =10-3=7

number of non-negative solution of above =$\binom{7+3-1}{3-1}$ =$\binom{9}{2}$ =9!/(2!*9!)=36