Matching problem

**tttcomrader** · June 10th 2006, 01:15 PM

My sister had this homework problem, and I'm not quite sure how to solve it.

There are 6 books that I want to buy, and I can only get 4 of them. How many different combination of books are there for me to choose from?

Is there a formula for this type of problem?

Thanks.

KK

**Soroban** · June 10th 2006, 02:40 PM

Hello, KK!

I hope your sister is familiar with factorials . . .

There are 6 books that I want to buy, and I can only get 4 of them.
How many different combination of books are there for me to choose from?

Yes, there is a formula . . .

This is a Combinations problem -- the order of the books is not important.

Choosing 4 items from 6, there are: $C(6,4) \:=\:\frac{6!}{4!2!}\:=\:15$ differemt choices.

**Soroban** · June 11th 2006, 07:35 AM

That is a "factorial" symbol.

Examples: . $\begin{Bmatrix}3!\:=\:1 \times 2 \times 3 \:=\:6 \\ 4!\:=\:1 \times 2 \times 3 \times 4\:=\:24 \\5! \:= \:1 \times 2 \times 3 \times 4\times 5 = 120 \\ \vdots \\ n!\:=\:1 \times 2 \times 3 \times \cdots \times n\end{Bmatrix}$

By defintion: . $1! = 1,\;\;0! = 1$

So: . $\frac{6!}{4!2!}\:=\:\frac{\not{1}\cdot\not{2}\cdot \not{3}\cdot\not{4}\cdot5\cdot\not{6}^3}{(\not{1} \cdot \not{2}\cdot\not{3}\cdot\not{4})(1\cdot\not{2})} \;= \;15$

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