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Solved Subjective Question on Permutations and Combinations Set 1

Posted on - 04-01-2017

Math

IIT JEE

Example.1

10 distinct balls have to be put in 3 different boxes. In how many ways can we do this if the order in which we put the balls in the boxes is also considered.

Solution

Let the balls be B1, B2,...,B10 and the boxes be b1, b2, b3.

Let us also introduce two identical markers X(say).

Now, if we take any permutations of 10 distinct balls and these two identical markers then these two markers divide the balls in 3 parts (in order) which we can put in the boxes b1, b2, b3, in the same order. So each permutation of these 12 objects in which 10 are distinct and 2 identical give a way of the requried arrangement. So number of required ways = No. of permutation of these 12 objects = .

Example.2

Suppose a man has 5 aunts and 6 uncles and his wife has 6 aunts and 5 uncles. How many way can he call a dinner party of 3 men and 3 women so that there are exactly 3 of the man’s relatives and 3 of the wife’s ?.

Solution

Looking at the set up, we have the following cases.

Man

Wife

3A

OU

3U

OA

(I)

2A

1U

1A

2U

(II)

2U

1A

2A

1U

(III)

3U

OA

3A

OU

(IV)

A and U denote aunt & uncle respectively.

Case I

5C3 6C0 5C3 6C0

100

Case II

5C2 6C1 6C1 5C2

3600

Case III

6C2 5C1 6C2 5C1

5625

Case IV

6C3 5C0 6C3 5C0

400

Total

9725

Example.3

In how many ways can the letters of the word CONCUBINE be arranged so that (a) the C's are never together (b) C's are always together.

Solution

(a) Ignoring the C's we have the letter ONUBINE

No. of ways to arrange these is 7!/2!, since there are two Ns here. .

Also, since the C's don't have to be together, we have the following arrangements:

X O X N X U X B X I X N X E X.

"C" can take the place of any of the positions marked X. Since, the number of such X positions is 8 and there are 2 C's, the number of ways we can select them is 8C2.

Hence the total number of ways is (7!/2!) 8C2.

(b) When the C's are together, we consider them as 1 combined unit, say CC. Thus, we have CC, O, U, B, I, N, E, N.

Number of ways we can arrange them is 8!/2! since there are 2 N’s again.

Example.4

A colour box has 3 red colours of different shades, 2 white colours of different shades and 7 green colours of different shades. In how many ways can 3 colours be taken from the box if at least one of them is red ?.

Solution

We have got 3 red colours, 2 white colours and 7 green colours.

There are 3 ways to take 3 colours from the box.

(1) 1red 2 other

(1) 2red 1 other

(1) 3red 0 other

(1) Number of ways to choose 1red = 3C1

Number. of ways to choose 2 others = 9C2.

(2) No. of ways to choose 2red = 3C2 .

No. of ways to choose 1 others = 9C1.

(3) No. of ways to choose 3red = 3C3 .

No. of ways to choose 0 others = 9C0.

Hence the total number of ways is

3C1 x 9C2 + 3C2 x9C1 + 3C3 x9C0 .

Example.5

Let n1 = x1 x2 x3 and n2 = y1y2 y3 be two 3 digit numbers. How many pairs of n1 and n2 can be formed so that n1 can be subtracted from n2 without borrowing.

Solution

Clearly n1 can be subtracted from n2 without borrowing if yi ≥ xi for i = 1, 2, 3 . Let xi = r, where r = 0 to 9 for i = 2 and 3.

and r = 1 to 9 for i = 1. .

Now as per our requirement yi = r, r +1,…. , 9. Thus we have (10 – r) choices for yi. .

Hence total ways of choosing yi and xi

= .

 
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