 # Combinations with Standard Results

Suppose someone asks you to draw four cards from a pack of 52 well-shuffled cards. You do so and note down the cards you drew. Again you shuffle the cards and draw another round of four cards. Are the cards drawn in both the cases the same? Maybe not, maybe they match partially or maybe they are the same. Here your interest does not lie in drawing the cards. It also does not interest you if in which draw which card was drawn. The real interest lies in the number of ways in which any four cards can be drawn. The number of ways in which we can select or draw any item of interest is the concept of the combination formula.

### Suggested Videos        ## Combination

The combination shows the number of the ways in which we can select different objects. In other words, each of the selections which can be made by taking some or all of the number of the things is a combination. In combination, the selection or the collection of things is irrespective of the order of selection.

When the selected things are arranged in all possible orders, each of the arrangements is a permutation. A combination for selecting r items from n items is denoted by the symbol nCr.

### Illustration

Suppose you have four fruits; Apple (A), Guava (G), Pear (P), and Orange (O). You want to select any two fruits from them. In how many ways can you select the two fruits? What are the possible selections? The possible selections are AG, AP, AO, GP, GO, and PO. The symbols have their respective meanings.

This selection of fruits does not depend upon the order of the selections. We can say that there are six possible ways of selection of two fruits out of four.

## Relation Between Permutation and Combination Formula

Consider the above example of selecting two fruits from the four. If we take into consideration the order in which we select a fruit first and then after the second one gives a situation of permutation. Also after selecting the fruits if we arrange them in an order the above situation of combination reduces to that of a permutation.

Since each of the two selected fruit can arrange themselves in 2! ways. We have in total of 12 ways of arranging the fruits. If there are n number of items out of which we need to select r of them. The relation between a permutation and a combination is

nPr = nCr × r!

or, n! ⁄ (n – r)! = nCr × r!

or, nCr = n! ⁄ (n – r)! r! ### Important Results Related to Combination Formula

• nCr  = nCn – r
• nCn = 1 = nC0 = 1 (0! = 1)
• If nCk = nCr, then either k = r or k + r = n.
• n + 1 Cr = nCr + nC r – 1.

## Combination Formula on Different Items

Let us consider the various situations of selecting r different items from n items when

### All Items are Different

Here, each item is categorized in two ways i.e., it may get included or excluded. Therefore, the total number of ways for either situation is 2 × 2 × 2 × 2 … n times = 2n. But this includes the case of rejection of all the items. So, the total number of ways of selection of one or more items = 2n – 1.

### All Items are Not Different

Suppose in a group of n items, there exist some objects which are of a similar kind and a few of them are different. Assume n = a + b + c + … + x. Out of these items, ‘a’ are of the first kind, ‘b’ are of the second kind, ‘c’ are of the third kind and so on and the remaining x are all different.

Out of ‘a’ items we can select 0, 1, 2, … , a items. Hence there are (a + 1) ways of selecting a. Similarly, the number of ways of selecting b is (b + 1) and so on. When the items are all different, the number of selecting x items is 2x. But this includes the case of rejection of all the items. So, the total number of ways of selections = (a + 1) (b + 1) (c + 1) 2x – 1.

## Solved Examples for You

Problem: Find the missing terms n and x.

1. nC8 = 17Cx.
2. nC6 = nC8.

Solution: We know that if nCp = nCq, then either p = q or p + q = n.

1. Given 8 + x = 17 or, x = 9. (8 cannot be equal to x as 8 + 8 ≠ 17).
2. We know, 6 + 8 = n = 14.

Problem: From a word containing 5 vowels and 12 consonants, how many 8 letter words can be formed by using 3 vowels and 5 consonants?

Solution: The number of ways of selecting 3 vowels from 5 vowels is 5C3. The number of ways of selecting 5 consonants from 12 consonants is 12C5. So, the number of ways of selecting 3 vowels and 5 consonants is  5C3 × 12C5. These eight letters can arrange themselves in 8! = 8P8 ways. Therefore, the total number of ways = 5C3 × 12C5 × 8! = 319334400 ways.

This concludes our disucssion on the topic of combination formula.

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