**High School Number Sense Lesson 52: Combinations**

On Tuesday we discussed the concept of

**Permutation**s. Today we will add to that discussion and introduce Combinations. This concept appeared

**13 times**last year, with a median placement at

**question # 54**.

**Number Dojo Level: 290**

A

**combination**is a way of selecting items from a set, but where the order doesn't matter. For example, if selecting 3 letters from the group

**{a,b,c,d}**, there are 4 different combinations:

- abc
- abd
- acd
- bcd

**Application:**

A common question on recent number sense tests has been "How many different committees of 3 people can be formed from a group of 5?" To solve this, we could spend our time counting them:

- 123
- 124
- 125
- 134
- 135
- 145
- 234
- 235
- 245
- 345

**Calculation:**

Or we could perform the calculation. Combinations are represented as "

**n**objects taken

**k**at a time," which is written as:

_{n}C_{k}**The calculation is performed by dividing**

**n!**by**k!(n - k)!**, or more visibly:**n!**

------------

k!(n - k)!

------------

k!(n - k)!

**Example 1: How many different committees of 3 people can be formed from a group of 5?**

- Divide
**5!**by**3!(5 - 3)!**, or**5!/(3!2!)** - Think of this as
**(5 x 4 x 3!)/(3!2!)**. Notice that the**3!**cancels from the top and bottom. - We are left with
**(5 x 4)/2!**, or 20/2, which is**10**.*Notice that this matches where we counted them above*.

**Example 2: The number of combinations of 6 objects taken 2 at a time is ___**

- Divide
**6!**by**2!(6 - 2)!**, or**6!/(2!4!)** - Think of this as
**(6 x 5 x 4!)/(2!4!)**. Notice that the**4!**cancels from the top and bottom. - We are left with
**(6 x 5)/2!**, or 30/2, which is**15**.

**Example 3: The number of combinations of 7 objects taken 4 at a time is ___**

- Divide
**7!**by**4!(7 - 4!)**, or**7!/(4!3!)** - Think of this as
**(7 x 6 x 5 x 4!)/(4!3!)**. Notice that the**4!**cancels from the top and bottom. - We are left with
**(7 x 6 x 5)/3!**, or**(7 x 6 x 5)/6**. Notice that the**6**cancels from the top and bottom. - We are left with
**(7 x 5)**, which is**35**.

**Example 4:**

_{8}C_{5}1. Divide

**8!**by

**5!(8 - 5)!**, or

**8!/(5!3!)**

2. Think of this as

**(8 x 7 x 6 x 5!)/(5!3!)**. Notice that the

**5!**cancels from the top and bottom.

3. We are left with

**(8 x 7 x 6)/3!**, or

**(8 x 7 x 6)/6**. Notice that the

**6**cancels from the top and bottom.

4. We are left with

**(8 x 7)**, which is

**56**.

**Example 5:**

**If**

_{5}C_{3}=_{5}C_{x}and x does not = 3, then x = ___1. This is more of a logic problem. The denominator of every Combination problem is

**k!(n - k!)**.

2. In this problem, n is 5 and k is 3. So the denominator would be

**3!(5 - 3)!**, or

**3!2!**.

3. The denominator would be the same if k were

**2**:

**2!(5 - 2)!**, or

**2!3!**. So

**2**is our answer.

**Example 6:**

_{7}P_{3}÷_{7}C_{3}1. This problem is MUCH easier than it looks. First we will solve it the long way.

2.

_{7}P

_{3}= 7!/(7 - 3)! This simplifies to 7!/4!, which is 7 x 6 x 5 =

**210**.

3.

_{7}C

_{3}= 7!/3!(7 - 3)! This simplifies to 7!/(3!4!), which is (7 x 6 x 5)/3!, which is 7 x 5 =

**35**.

4. 210 ÷ 35 =

**6**.

5.

**The trick:**No matter what the numbers are, the factorials will all cancel out except for

**k!**. In this problem, k = 3, and 3! is

**6**, which was our answer.

**Here's a free worksheet to help you practice Combination:**

combination.pdf |

**Up Next for High School: ComplexNumber**