Number of Combination is the number of selections that can be made for a given number of objects. it is interesting to note that the number of combinations is always lesser than the number of permutations. The number of combinations is the total number of ways in which you can select a given set from a larger set. In combinations, the order of selection doesn’t matter. Otherwise, they would become the number of permutations. Here we will see examples for counting the combinations of a set or a collection of objects. Let us see!

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## Number of Combination

Let us say that we have a group of four students. We label one of the students as ‘a’, the others as ‘b’, ‘c’ and ‘d’ respectively. How many ways can you select a group of two out of these four people? Let us see how to utilise the concept of combinations to make the process of counting selections easier. Here say we may have the following selections of sets: {a, b} , {b, c}, { c, d} … But writing or counting the number of selections of combinations like this will be impossible as the number of parent group gets larger and larger. There must be an easier way!

We have already seen what permutations are. The number of different ways in which we can arrange the elements of a set can be found out by counting the number of permutations. We also know that the number of permutations, when divided by the number of arrangements, got by neglecting order gives the number of combinations. Mathematically we can write that if a selection has to be made out of a set of ‘n’ objects taking ‘r’ objects at a time, then the number of combinations will be given by ^{n}C_{r} = (n!)/r!(n-r)!

### Some Examples Of The Number Of Combinations

Let us see some solved examples that will make us familiar with the formula and the concept of counting the number of selections.

Example 1: The value of ^{n}C0_{ }and ^{n}C_{n} have a ratio of:

A) 1: 1 B) n:1 C) 1:n D) 1:2

Answer: Let us first calculate the value of ^{n}C0 = (n!)/0!(n-0)!. Since we know that 0! = 1, we can write:

^{n}C0 = n!/n! = 1. Now let us find the value of ^{n}Cn = n!/n!(n-n)! = n!/n!(0)! = 1.

Thus the correct option is A) 1:1 or that ^{n}C0_{ }= ^{n}C_{n}

What this means is that the number of selections that are possible for a group of ‘n’ elements taken none at a time is equal to the number of possible selections that we can make taking all at a time = 1. In the first case, we don’t select anything and that counts as one selection!

## Solved Examples

Example 2: In how many ways can you select a four-letter word in the English alphabet?

Answer: The group that we have to make our selection from gives us the value of ‘n’. Here the group is the English alphabet and hence n = 26. The number of elements or objects that we group together at a time gives us the value of ‘r’. So the number of selections that can be done for this four-letter word is found by ^{26}C_{4}.

Let us add to the example a bit. What is the total number of four-letter words that can be formed from the English alphabet? Will it be equal to ^{26}C_{4}? The answer is no. This is obvious once you notice that when we are performing selections, the order doesn’t matter. For example, if you select A and then B, it is equivalent to selecting B and then A. So we say that for “selections” order of the selection is immaterial. But when you are forming words the order becomes important. This is the number of arrangements or the number of permutations. They are always greater than the number of combinations. Let us see more examples!

### Type II

Example 3: A committee of 5 people is to be chosen from a group of 6 men and 4 women. How many committees are possible if there are to be 3 men and 2 women in each committee?

Answer: Split the problem into smaller sections. here we have two groups of people to choose from. One is the group of males and the other the group of females. Therefore, for the first group, we have to choose 3 men out of 6 men. This can be done in ^{6}C_{3} ways. Similarly, from the females, we have to choose 2 women from a group of 4 women. That means we have to make ^{4}C_{2} selections.

Therefore the total number of selections that we can make will be the multiplication of the two. In other words, we can say the total selections are = ^{6}C_{3} × ^{4}C_{2} = 120. So there are 120 ways to select the committee. How many committees will be there? Since in a committee, the order doesn’t matter. As long as a person is on the committee, it doesn’t matter if he or she was selected first or second. So here the number of combinations will give the number of committees as order doesn’t matter.

## Practice Problems:

Q 1: If 4 Maths books are selected from 6 different Maths books and 3 English books are chosen from 5 different English books, how many ways can the seven books be arranged on a shelf if there are no restrictions?

A) 756000 B) 124 C) 1488 D) 9087456

Ans: A) 756000

Q 2: In a hand of poker, 5 cards are dealt from a regular pack of 52 cards. In how many of these hands are there exactly 4 Kings?

A) 58 B) 48 C) 68 D) 986

Ans: B) 48

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