In how many ways can 8 books be chosen from a group of nine

A committee of 5 is to be chosen from a group of 9 people.Number of ways in which it can be formed if two particular persons either serve together or not at all and two other particular persons refuse to serve with each other.

Let $P_1,P_2$ be the persons who either serve together or not at all and $P_3,P_4$ be the persons who refuse to serve with each other. Possible cases are $(P_1,P_2,P_3)$ and two others,$(P_1,P_2,P_4)$ and two others,$P_3$ and 4 others excluding $P_1,P_2,P_4$ and $P_4$ and 4 others excluding $P_1,P_2,P_3.$

Total cases are$=5C2+5C2+5C4+5C4=30$ but the answer is $41$

Learning Outcomes

• Find the number of combinations of n distinct choices.

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 [latex]r[/latex] objects from a set of [latex]n[/latex] objects where the order does not matter can be written as [latex]C\left(n,r\right)[/latex]. Just as with permutations, [latex]\text{C}\left(n,r\right)[/latex] can also be written as [latex]{}_{n}{C}_{r}[/latex]. In this case, the general formula is as follows.

[latex]\text{C}\left(n,r\right)=\dfrac{n!}{r!\left(n-r\right)!}[/latex]

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 [latex]3!=3\cdot 2\cdot 1=6[/latex] ways to order 3 paintings. There are [latex]\frac{24}{6}[/latex], 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 [latex]n[/latex] distinct objects, the number of ways to select [latex]r[/latex] objects from the set is

[latex]\text{C}\left(n,r\right)=\dfrac{n!}{r!\left(n-r\right)!}[/latex]

How To: Given a number of options, determine the possible number of combinations.

1. Identify [latex]n[/latex] from the given information.
2. Identify [latex]r[/latex] from the given information.
3. Replace [latex]n[/latex] and [latex]r[/latex] in the formula with the given values.
4. Evaluate.

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?

No. When we choose r objects from n objects, we are not choosing [latex]\left(n-r\right)[/latex] objects. Therefore, [latex]C\left(n,r\right)=C\left(n,n-r\right)[/latex].

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

We have looked only at combination problems in which we chose exactly [latex]r[/latex] 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 [latex]C\left(5,0\right)=1[/latex] way to order a pizza with no toppings. There are [latex]C\left(5,1\right)=5[/latex] ways to order a pizza with exactly one topping. If we continue this process, we get

[latex]C\left(5,0\right)+C\left(5,1\right)+C\left(5,2\right)+C\left(5,3\right)+C\left(5,4\right)+C\left(5,5\right)=32[/latex]

There are 32 possible pizzas. This result is equal to [latex]{2}^{5}[/latex].

We are presented with a sequence of choices. For each of the [latex]n[/latex] objects we have two choices: include it in the subset or not. So for the whole subset we have made [latex]n[/latex] choices, each with two options. So there are a total of [latex]2\cdot 2\cdot 2\cdot \dots \cdot 2[/latex] possible resulting subsets, all the way from the empty subset, which we obtain when we say “no” each time, to the original set itself, which we obtain when we say “yes” each time.

A set containing n distinct objects has [latex]{2}^{n}[/latex] subsets.

A restaurant offers butter, cheese, chives, and sour cream as toppings for a baked potato. How many different ways are there to order a potato?

A sundae bar at a wedding has 6 toppings to choose from. Any number of toppings can be chosen. How many different sundaes are possible?

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It depends. Does the order of the group of 4 people matter? Let's say your group of 9 consists of Alice, Bob, Carly, David, Ernest, Fran, Gillian, Henry, and Iris. Clearly, Alice, Bob, Carly, and David is one set of 4 people that you could select. The question is this: Is Alice, Bob, Carly, and David different from David, Carly, Bob, and Alice? If you were just selecting 4 committee members, then the order doesn't matter, but if you were selecting a Club President, Vice-President, Secretary, and Treasurer, then order would definitely matter.

So, if order matters, then you have a permutation of 9 things taken 4 at a time given by:

If order does not matter, then you have a combination, and the answer you got for the permutation is too large by a factor of the number of ways that you can arrange the group of 4, namely 4!. So you need to divide the permutation calculation by that amount: The combination of 9 things taken 4 at a time is then: