# Solving Combination Problem: 11Cn=330

• temaire
In summary, to solve for n in the equation 11_{C}_n=330, you can use the combination formula n_{C}_r=\frac{n!}{(n-r)!r!} and algebra to reduce the equation to n!(11-n)!=3 \cdot 8!. Then, by guessing and checking, you can find that n can either be 4 or 7 to satisfy the equation.

## Homework Statement

Solve for n.
$$11_{C}_n=330$$

## Homework Equations

$$n_{C}_r=\frac{n!}{(n-r)!r!}$$
Sorry if the combination formula looks bad. I don't know how to write the comb. formula with Latex.

## The Attempt at a Solution

I solved for n as far as $$n!(11-n)!=133056$$ How do I go further with this? The answers are 4 or 7.

Honestly the easiest way to do this would be a combination of algebra and guess and check. You should see that both n and 11-n would solve the equation, so finding one answer would give you the answer.

The usual way to reduce factorials is to try to keep as much of the factorial together as possible, which often let's you see patterns easier. For instance:

$$_{11}C_n = \frac{11!}{n!(11-n)!} = 330 \implies n!(11-n)! = \frac{11!}{2 \cdot 3 \cdot 5 \cdot 11}$$

You can cancel out the 11 and the 10 on top, leaving you with:

$$n!(11-n)! = 3 \cdot 8!$$

Now you start with the guess and check. You know that n has to be less than 8.

I would approach this problem by first understanding the concept of combinations and how they are related to factorials. The combination formula, as you mentioned, is nCr = n! / (r!(n-r)!), where n represents the total number of items and r represents the number of items being chosen at a time. In this case, we are given that n = 11 and nCr = 330.

To solve for n, we can rearrange the formula to get n! = 330(r!(11-r)!). We can then try different values for r and see if the resulting equation is satisfied. For example, if we let r = 4, we get n! = 330(4!(11-4)!) = 330(4x7!) = 330x5040 = 166320. This is not equal to 330, so we can try another value for r.

If we let r = 7, we get n! = 330(7!(11-7)!) = 330(7x4!) = 330x5040 = 166320. This is equal to 330, so we have found a valid solution. Thus, n = 11 and r = 7.

In conclusion, the solution to this combination problem is n = 11 and r = 7. This can be verified by plugging these values back into the combination formula to get 11C7 = 330.

## 1. What is a combination problem?

A combination problem is a type of mathematical problem where the order of the items does not matter. It involves selecting a certain number of items from a larger set of items.

## 2. How do you solve a combination problem?

To solve a combination problem, you can use the formula nCr = n! / r!(n-r)!, where n represents the total number of items and r represents the number of items being selected. You can also use a combination calculator or a tree diagram to find the solution.

## 3. What is the value of n in the equation 11Cn=330?

In this equation, n represents the number of items being selected from a set of 11 items. It is the variable that we need to solve for in order to find the solution to the combination problem.

## 4. How do you find the value of n in the equation 11Cn=330?

To find the value of n, we can rearrange the equation to solve for n. By using the formula nCr = n! / r!(n-r)!, we can rewrite the equation as n! / (n-11)!11! = 330. Then, using trial and error or a combination calculator, we can determine that n = 15.

## 5. What is the solution to the combination problem 11Cn=330?

The solution to this combination problem is n = 15. This means that there are 15 different combinations of 11 items that can be selected, with the total number of combinations being 330.