The symbol ^{n}C_{r }(pronounced n choose r) denotes the number of ways in which we can choose ‘r’ objects from a collection of ‘n’ number of objects. The combination formula is used in statistics and probability to calculate the number of possibilities of an outcome.

**Example**: Suppose you have three people A, B, and C out of whom you want to choose two of them to be part of a tennis team.

Then we have the following possibilities for the team:

- A and B are chosen as the players for the team.
- A and C are chosen as the players for the team.
- B and C are chosen as the players for the team.

So, we see that there are three different ways to choose 2 people out of 3. Hence we conclude that ^{3}C_{2} = 3. Also notice that in the above example, the order in which the people were chosen did not matter. Our tennis team would have been the same regardless of whether player A was chosen first followed by player B OR player B was chosen first followed by player A. This is true in general for the combination formula.

*The combination formula does not take into account the order in which the objects are chosen from the bigger collection.* The different re-orderings of objects that are possible are considered to be the same for the purpose of the combination formula.

**Formula to calculate the number of Combinations:**

The combination formula is given as,

Here, n! = product of all natural numbers less than equal to n (*Example*: 5! = 5x4x3x2x1 = 120)

So if we want to calculate ^{3}C_{2} the formula tells us that,

^{3}C_{2} = 3!/(2!x1!) = (3x2x1)/(2x1x1) = 6/2 = 3 which agrees with the answer in the above example.

**Example 1**: A bag contains 5 balls. Calculate the number of ways in which we can choose three balls from the bag.

**Solution**: We need to calculate ^{5}C_{3} (pronounced 5 choose 3). Applying the above formula we get,

^{5}C_{3} = 5!/(3!x2!) = (5x4x3x2x1)/(3x2x1x2x1) = 120/12 = 10

So, there are 10 different ways to choose 3 balls out of 5.

*Shortcut to calculate ^{5}C_{3} (5 choose 3)*: We can shorten the above calculation by the following trick. Instead of multiplying the numerator and denominator separately and then dividing the result, we can first cancel the common terms occurring in both the numerator and denominator in 5! And 3! and then simplify the fraction. So we see that 3,2 and 1 get canceled in the calculation below,

So, ^{5}C_{3} = 5!/(3!x2!) = (5x4x~~3~~x~~2~~x~~1~~)/(~~3~~x~~2~~x~~1~~x2x1) = (5×4)/(2×1) = 20/2 = 10

This shortcut can be applied in general to calculate ^{n}C_{r }(n choose r). We can simplify the calculation by striking the common factors between n! and r!.

**Example 2**: Calculate the number of ways in which we can choose 2 balls from a bag containing 6 balls.

**Solution**: We calculate ^{6}C_{2} (6 choose 2) as follows,

^{6}C_{2} = 6!/(4!x2!) = (6x5x4x3x2x1)/(4x3x2x1x2x1) = (6×5)/(2×1) = 30/2 =15

So there are 15 different ways to choose 2 balls out of 6.