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Introduction to Combination

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Introduction to Combination - Lesson Summary

The notion of permutation is based on the 'order' of the arrangement of objects.

Order of the arrangement is not concerned in arriving at the concept of combinations.

Consider three points on a plane.

To find the number of straight lines that can be drawn through these three points-

Taking two points at a time, we can draw three straight lines, AB, BC and CA.

Though a straight line can be drawn from B to A, it is considered equivalent to the straight line from A to B. It implies that AB and BA are equal.

The order in which the points are chosen to draw a line is not important.

A similar analogy can be drawn for the lines drawn from C to B, and from A to C.

We get three straight lines joining points A, B and C.

We can say that the straight lines are drawn taking a combination of two points at a time.

Formula for computing combinations

Take three objects, X, Y and Z.

Form combinations of two objects using the given objects.

The combinations are XY, XZ , YZ.

The total number of combinations is 3.

Using 3 different objects, taken 2 at a time, we get 3 distinct combinations.

This can be expressed mathematically as 3C2 = 3.

Possible permutations using these objects taken two at a time are XY, YX, XZ, ZX, YZ, ZY.

The number of permutations is six.

3P2 = 6

Each combination has two permutations.




Each combination of two letters can be permutated in two ways or two factorial ways.

Therefore, by the multiplication rule, each combination multiplied by two factorial ways is equal to the permutation.

3C2 x 2! = 3P2

3C2 x 2! = 3 x 2! = 6 = 3P2

This relation can be generalised for any number of objects.

nCr x r! = nPr

⇒( nCr x r! )/r! = nPr/r!

nCr = nPr/r!

nCr = n!/((n-r)!r!) [Number of combinations of n objects taken r at a time]

Note: The value of 0! is taken as 1.

Example: Compute the number of selections possible for a volleyball team of six members that has to be selected out of eight members.

Six members out of eight members are selected.

n = 8 r = 6

8C6 = 8!/((8-6)!6!) = 28

This implies that the team can be selected in 28 different ways.


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