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Mathematical Methods - Combinations | 11th Statistics : Chapter 7 : Mathematical Methods

Chapter: 11th Statistics : Chapter 7 : Mathematical Methods

Combinations

Combination is a selection of objects without considering the order of arrangements.

Combinations

Combination is a selection of objects without considering the order of arrangements. For example out of three things A, B, C we have to select two things at a time. This can be selected in three different ways as follows.

 

AB AC BC

 

Here the selection of object AB and BA are one and the same. The order of arrangement is not considered in combination. Hence the number of combinations from 3 different things taken 2 at a time is 3.

This is written symbolically 3C2  = 3.

Now we use the formula to find combination.

The number of combination of n different things, taken r at a time is given by


 

Example 7.8

Find 10C3  and 8C4

Solution:


To find 10 C 3 : In the numerator, first write the product of 3 natural numbers starting from 10 in the descending order and in the denominator write the factorial 3 and then simplify ].


 

Example 7.9

Out of 13 players, 11 Players are to be selected for a cricket team. In how many ways can this be done?

Solution:

Out of 13 Players, 11 Players are selected in 13C11 ways


 

Example 7.10

In how many ways a committee of 5 members can be selected from 6 men and 5 women, consisting of 2 men and 3 women?

Solution:

For a committee, 2 men and 3 women members are to be selected. From 6 men, 2 men are selected in 6C2 ways. From 5 women, 3 women are selected in 5C3 ways.

Hence a committee of 5 members (2 men and 3 women) is selected in


 

Example 7.11

How many triangles can be formed by joining the vertices of a pentagon of five sides.

Solution:

There are 5 vertices in a pentagon. One triangle is formed by selecting a group of 3 vertices from given 5 vertices. This can be done in 5C3 ways.


 

Example 7.12

A question paper contains section A with 5 questions and section B with 7 questions.

A student is required to attempt 8 questions in all, selecting at least 3 from each section.

In how many ways can a student select the questions?

Solution:

Selection of 8 questions from 12 questions and at least 3 from each section is given below.


Therefore total number of selection is 420

 

Example 7.13

If 6 Pr   = 360 and 6Cr = 15 find r.

Solution:

From the formula,


 

Example 7.14

If nC8  = n C7,   find n C15

Solution:

nC8 = n C7

 nCn–8 = nC7

 n–8 = 7

 n = 15

 Now, nC15 = 15C15 = 1.

 

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