13.5 Counting principles  (Page 4/12)

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Find the number of combinations using the formula

So far, we have looked at problems asking us to put objects in order. There are many problems in which we want to select a few objects from a group of objects, but we do not care about the order. When we are selecting objects and the order does not matter, we are dealing with combinations . A selection of $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ objects from a set of $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ objects where the order does not matter can be written as $\text{\hspace{0.17em}}C\left(n,r\right).\text{\hspace{0.17em}}$ Just as with permutations, $\text{\hspace{0.17em}}\text{C}\left(n,r\right)\text{\hspace{0.17em}}$ can also be written as ${\text{\hspace{0.17em}}}_{n}{C}_{r}.\text{\hspace{0.17em}}$ In this case, the general formula is as follows.

$\text{\hspace{0.17em}}\text{C}\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}\text{\hspace{0.17em}}$

An earlier problem considered choosing 3 of 4 possible paintings to hang on a wall. We found that there were 24 ways to select 3 of the 4 paintings in order. But what if we did not care about the order? We would expect a smaller number because selecting paintings 1, 2, 3 would be the same as selecting paintings 2, 3, 1. To find the number of ways to select 3 of the 4 paintings, disregarding the order of the paintings, divide the number of permutations by the number of ways to order 3 paintings. There are $3!=3·2·1=6$ ways to order 3 paintings. There are $\frac{24}{6},\text{\hspace{0.17em}}$ or 4 ways to select 3 of the 4 paintings. This number makes sense because every time we are selecting 3 paintings, we are not selecting 1 painting. There are 4 paintings we could choose not to select, so there are 4 ways to select 3 of the 4 paintings.

Formula for combinations of n Distinct objects

Given $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ distinct objects, the number of ways to select $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ objects from the set is

$\text{\hspace{0.17em}}\text{C}\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}\text{\hspace{0.17em}}$

Given a number of options, determine the possible number of combinations.

1. Identify $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ from the given information.
2. Identify $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ from the given information.
3. Replace $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}r\text{\hspace{0.17em}}$ in the formula with the given values.
4. Evaluate.

Finding the number of combinations using the formula

A fast food restaurant offers five side dish options. Your meal comes with two side dishes.

1. How many ways can you select your side dishes?
2. How many ways can you select 3 side dishes?
1. We want to choose 2 side dishes from 5 options.
$\text{\hspace{0.17em}}\text{C}\left(5,2\right)=\frac{5!}{2!\left(5-2\right)!}=10\text{\hspace{0.17em}}$
2. We want to choose 3 side dishes from 5 options.
$\text{\hspace{0.17em}}\text{C}\left(5,3\right)=\frac{5!}{3!\left(5-3\right)!}=10\text{\hspace{0.17em}}$

Is it a coincidence that parts (a) and (b) in [link] have the same answers?

No. When we choose r objects from n objects, we are not choosing $\text{\hspace{0.17em}}\left(n–r\right)\text{\hspace{0.17em}}$ objects. Therefore, $\text{\hspace{0.17em}}C\left(n,r\right)=C\left(n,n–r\right).\text{\hspace{0.17em}}$

An ice cream shop offers 10 flavors of ice cream. How many ways are there to choose 3 flavors for a banana split?

$\text{\hspace{0.17em}}C\left(10,3\right)=120\text{\hspace{0.17em}}$

Finding the number of subsets of a set

We have looked only at combination problems in which we chose exactly $r$ objects. In some problems, we want to consider choosing every possible number of objects. Consider, for example, a pizza restaurant that offers 5 toppings. Any number of toppings can be ordered. How many different pizzas are possible?

To answer this question, we need to consider pizzas with any number of toppings. There is $C\left(5,0\right)=1$ way to order a pizza with no toppings. There are $C\left(5,1\right)=5$ ways to order a pizza with exactly one topping. If we continue this process, we get

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