Question

a bookstore sells 5 types of notebooks. if a student buys 3 notebooks in total, and the order doesnt matter and they can buy multiple of the same type, how many different selections are possible?
answer =

how many solutions solutions does the equation w+x+y+z= 5 have, where w, x, y, and z are nonnegative integers (i.e. at least 0)?
answer =

Explanation:

Step1: Identify combination formula (repetition)

This is a stars and bars problem. The formula for combinations with repetition is $\binom{n+k-1}{k}$, where $n$ = number of types, $k$ = number of items to choose.

Step2: Solve first notebook problem

$n=5$, $k=3$. Calculate $\binom{5+3-1}{3} = \binom{7}{3}$
$\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7\times6\times5}{3\times2\times1} = 35$

Step3: Identify equation solution formula

For non-negative integer solutions to $w+x+y+z=5$, use $\binom{n+k-1}{k}$, where $n=4$ variables, $k=5$.

Step4: Solve equation solution problem

Calculate $\binom{4+5-1}{5} = \binom{8}{5}$
$\binom{8}{5} = \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8\times7\times6}{3\times2\times1} = 56$

Answer:

1. 35
2. 56