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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 $\text{\hspace{0.17em}}C(n,r).\text{\hspace{0.17em}}$ Just as with permutations, $\text{\hspace{0.17em}}\text{C}(n,r)\text{\hspace{0.17em}}$ 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.
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 $3!=3\xb72\xb71=6$ 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.
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
Given a number of options, determine the possible number of combinations.
A fast food restaurant offers five side dish options. Your meal comes with two side dishes.
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 $\text{\hspace{0.17em}}(n\u2013r)\text{\hspace{0.17em}}$ objects. Therefore, $\text{\hspace{0.17em}}C(n,r)=C(n,n\u2013r).\text{\hspace{0.17em}}$
An ice cream shop offers 10 flavors of ice cream. How many ways are there to choose 3 flavors for a banana split?
$\text{\hspace{0.17em}}C(10,3)=120\text{\hspace{0.17em}}$
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 $C(5,0)=1$ way to order a pizza with no toppings. There are $C(5,1)=5$ ways to order a pizza with exactly one topping. If we continue this process, we get
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