Joe has one book each for algebra, geometry, history, psychology, spanish, english and physics in his locker. how many different sets of three books could he choose?

There are 35 different sets of 3 books Joe could choose

Step-by-step explanation:

* Lets explain how to solve the problem

- Combination is a collection of the objects where the order doesn't

 matter

- The formula for the number of possible combinations of r objects from

 a set of n objects is nCr = n!/r!(n-r)!

- n! = n(n - 1)(n - 2)................. × 1

Lets solve the problem

- Joe has one book each for algebra, geometry, history, psychology,

 Spanish, English and Physics in his locker

∴ He has seven books in the locker

- He wants to chose three of them

∵ The order is not important when he chose the books

∴ We will use the combination nCr to find how many different sets

  of three books he can choose

- The total number of books is 7

∴ n = 7

∵ He chooses 3 of them

∴ r = 3

∵ 7C3 = 7!/3!(7 - 3)! = 7!/3!(4!)

∴

∴ 7C3 = 35

* There are 35 different sets of 3 books Joe could choose

lol we need to see what the question is asking about

step-by-step explanation:

there is only one way to arrange all the eight relative to each other because they are not distinct. we have

now there are nine places to place an (and not more than one in one place).   because we have four symbols, we need to choose four of these nine places to insert an . order does not matter because the are not distinct.

thus there are 1 · c(9,4) ways, or, if we calculate it,

thus there are 126 ways to form such "words."