Permutation and Combination (Exercise 1.3)

1. A boy puts his hand into a bag which contains 10 differently coloured marbles and brings out 3. How many different results are possible?
Solution:  3 marbles can be choosen from 10 marbles in,
                C(10, 3) = $\frac{10!}{7!.3!}$ = $\frac{10\ast <!--\ast -->9\ast <!--\ast -->8}{3\ast <!--\ast -->2\ast <!--\ast -->1}$ = 120 ways.  



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Find the number of ways in which a student can select 5 courses out of 8 courses. If 3 courses are compulsory, in how many ways can the selections be made?

Solution: The student can be select 5 courses out of 8 courses in,
                C(8, 5) = $\frac{8!}{\left(8-<!----->5\right)!5!}$ = $\frac{8\ast <!--\ast -->7\ast <!--\ast -->6\ast <!--\ast -->5!}{3\ast <!--\ast -->2\ast <!--\ast -->1\ast <!--\ast -->5!}$ = 56 ways.
Out of 8 courses 3 courses are compulsory therefore there is selection = 5 - 3 = 2 courses out of if 8 - 3 = 5 courses is C(5, 2)
$\frac{5!}{\left(5-<!--\; -\; -->2\right)!.12!}$ = $\frac{5\ast <!--\; \ast \; -->4\ast <!--\; \ast \; -->3\ast <!--\; \ast \; -->2\ast <!--\; \ast \; -->1}{3\ast <!--\; \ast \; -->2\ast <!--\; \ast \; -->1\ast <!--\; \ast \; -->2\ast <!--\; \ast \; -->1}$ = 10 ways.



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From 10 persons in how many ways can a selection of 4 be made,

1. When one particular person is always included?
2. When two particular persons are always excluded?
Solution:
1. When one particular person is included, 3 persons have to be selected from remaining 9 persons.
   This can be selected in, C(9, 3) =
$\frac{9!}{6!.3!}$ = $\frac{9\ast <!--\; \ast \; -->8\ast <!--\; \ast \; -->7}{3\ast <!--\; \ast \; -->2\ast <!--\; \ast \; -->1}$ = 84 ways.

2. When two particular persons always excluded, 4 persons have to be selected from remaining 8 persons. This can be selected in,
   C(8, 4) = $\frac{8!}{4!.4!}$ = $\frac{8\ast <!--\; \ast \; -->7\ast <!--\; \ast \; -->6\ast <!--\; \ast \; -->5}{4\ast <!--\; \ast \; -->3\ast <!--\; \ast \; -->2}$ = 70 ways.



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2. A bag contains 8 white balls and 5 blue balls. In how many ways can 5 white balls and 3 blue balls be drawn?
Solution: 5 white balls can be drawn from 8 white balls in C(8, 5) different ways,
3 blue balls can be drawn from 5 blue balls in C(5, 3) different ways.
Therefore, total number of selection (drawn) is,
C(8,5)*C(5,3) = $\frac{8!}{\left(8-<!--\; -\; -->5\right)!.5!}$ * $\frac{5!}{\left(5-<!--\; -\; -->3\right)!.3!}$ = 56 * 10 = 560 ways.


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