Let us suppose there are four persons A, B, C and D (actual names may be used here) and we have to select three of them to be a part of a committee.

**Combinations**

Let us suppose there
are four persons *A, B, C* and *D* (actual names may be used here) and we have to select three of
them to be a part of a committee. In how many ways can we make this selection?
For example, *A, B, C* is one possible
choice. Here the order of selection is immaterial. Thus *A, B, C* is the same as *B, A, C* or *C, A, B* as long as the same three persons are selected. Thus the
possible distinct choices or selections are *A, B, C*; *A, B, D*; *A, C, D* and *B, C, D*. We may thus conclude that there are 4 ways of selecting 3
people out of 4. Each choice or selection is referred to as a combination of 4
different objects taken 3 at a time.

Suppose two persons
are to be selected from four persons. The possible choices are: *A, B* : *A, C*: *A, D *:* B, C *:* B, D *:* C, D*. Thus the number of combinations of 4
different objects taken* *2* *at
a time* *is 6. The number of
combinations of *n* different objects
taken *r* at a time is represented by ^{n}*C _{r}*. From the above we may conclude that

Thus the total number
of permutations of 4 objects taken 3 at a time is ^{4}*C*_{3} *×* 3!. This is also equal
to ^{4}*P*_{3}. Hence* *^{4}*P*_{3}* *=^{4}* C*_{3}* × *3!*.*

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11th Mathematics : UNIT 4 : Combinatorics and Mathematical Induction : Combinations | Definition, Formula, Solved Example Problems, Exercise | Mathematics

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