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Solved Subjective Question on Permutations and Combinations Set 2

Posted on - 04-01-2017

Math

IIT JEE

Example.1

(a). In how many ways can 7 men and 7 women be seated around a circular table so that no two men/no two women sit next to each other ?.

(b). Suppose that at a sports dinner we have 16 cricketers and 6 tennis players. In how many ways can we seat them at a long table if (i) none of the tennis players is seated next to another tennis player and (ii) all tennis players are seated together.

(c). There are 20 persons among whom are two brothers. Find the number of ways in which we can arrange them around a circle so that there is exactly one person between the two brothers.

(d). A tea party is arranged for 2m people along two sides of a long table with m chairs on each side. r men wish to sit on one particular side and s on the other. In how many ways can they be seated? (Assume that r, s £ m.).

(e). A family consists of a grandfather, m sons and daughters and 2n grandchildren. They are to be seated in a row for dinner. The grandchildren wish to occupy the n seats at each end and the grandfather refuses to have a grandchild on either side of him. In how many ways can the family be made to sit?.

Solution

(a). First arrange the 7 men in 6! ways. Then there are 7 places in between the men which will be occupied by 7 women in 7!, ways. So total no. of ways = 6! ´ 7!.

(b). (i) First, 16 cricketers can be seated in 16! ways. There are 17 places in between them out of which 6 have to be occupied by tennis players.
No
. of ways = 17P6 .

\ Required no. of ways = 16! ´ 17P6 .

(ii) All tennis players can be treated as one set. So now we have a total of 17 players. Their no. of permutations = 17 !.

Also 6 tennis players can be arranged in 6! ways. .

No. of ways = 17 ! ´ 6!.

(c) We can arrange 18 persons around a circle in (18 –1)! = 17! ways. Now, there are exactly 18 places where we can arrange the two brothers. Also the two brothers can be arranged in 2! ways. Thus, the number of ways of arranging the persons subject to the given condition is (17!) (18) (2!) = 2(18!).

(d) We can arrange r persons on m chairs on a particular side in mPr ways and s persons on m chairs on the other side in mPs ways. We can arrange (2m –r –s ) persons on the remaining (2m –r –s ) chairs in 2m –r –s P2m –r –s ways. Thus, the number of ways of arranging the persons subject to the given conditions is (mPr) (mPs) (2m –r –s P2m –r –s).

(e) The total number of seats required at the table is 1 + m + 2n. The grand children can occupy the n seats on either side of the table in (2nP2n) ways. The grandfather can occupy a seat in m –1P1 ways. The remaining seats can be occupied in mPm ways. Therefore, the required numbers of ways is.

(2nP2n) (mPm) (m –1 P1) = (2n!) m! (m –1)

Example.2

There are five balls of different colours and five boxes of the same colours as those of the balls. Find the number of ways in which the balls could be placed one each in a box, so that no ball goes to the box of its own colour. .

Solution

Let us denotes the colours by 1, 2, 3, 4 and 5. Let Ai denotes the set of distributions of balls in the boxes so that with ith colour goes to the box of its own colour.

We have n( Ai) = 4! " i,

n(Ai Ç Aj) = 3! ( i > j) ,

n( Ai Ç Aj ÇAk) = 2! ( i, j, k are distinct ), etc.

We have n(A1È A2ÈA3ÈA4ÈA5)

=

= 5 (4!) – 5C2(3!) +5C3(2!) – 5C4 (1!) +5C5

=

Thus, the number of ways in which no ball goes into the box of its own colour is 5! –

= 60 – 20 + 5 – 1= 44.

Example.3

Six apples and six mangoes are to be distributed among ten boys so that each boy receives at least one fruit. Find the number of ways in which the fruits can distributed. .

Solution

When 12 fruits are distributed subjected to the given condition, either nine boys get one fruit each and the remaining one boy gets three fruits or eight boys get one fruit each and the remaining two boys get two fruit each. Three fruits to a boy can be given in the four ways.

Apples

3

2

1

0

Mangoes

0

1

2

3

After giving 3 apples to a single boy, the 3 remaining apples and 6 oranges can be distributed to the 9 boys in (9C3) (6C6) ways.

Thus, three fruits can be given to a particular boy in

One particular boy can be chosen in 10C1 = 10 ways

Therefore, three fruits can be given to a single boy in

(10C1) (2) [9C3+9C4] = 20 (10C4) = 4200 ways

We can give two fruits to two boys, say P and Q in the following ways.

P

Apples

2

1

0

2

1

0

2

1

0

Mangoes

0

1

2

0

1

2

0

1

2

Q

Apples

2

2

2

1

1

1

0

0

0

Mangoes

0

0

0

1

1

1

2

2

2

The remaining eight fruits can be distributed among eight boys in the following ways:

+

+ =

Two boys out of 10 can be chosen in 10C2 ways.

Therefore, two boys can get two fruits each in

=

Hence, the required number of ways = 4200 + 22050 = 26250.

Example.4

Find the number of natural numbers which are less than 2 ´ 108 and are divisible by 3 and which can be written by means of the digits
0, 1, 2
. .

Solution

A natural number which is smaller than 2.108 and which can be written by means of the digits 0, 1 and 2 is of the form a1 a2 … a9 where 0 £ a1 £ 1 and 0 £ ai £ 2 for i = 2, 3, ….,9 and where all a1, a2, …, a9 cannot be equal to zero.

We can choose a1 in two ways (0 or 1) and ai for i = 2, 3,…, 8 in three ways (0, 1 or 2). After choosing a1, a2, …. , a8 we find the sum s = a1 + a2 + …+ a8 is of the form 3m – 2, 3m – 1 or 3m. We can now choose a9 in just one way. Infact a9 = 2, 1 or 0 depending on whether S = 3m – 2, 3m – 1 or 3m. Therefore, we can choose the numbers in (2) (37) (1) ways. But this includes a choice in which each of ai = 0. Thus, the required number of numbers = (2) (37) (1) – 1 = 4373.

Example.5

In an examination, the maximum marks for each of the three papers are 50 each. Maximum marks for the fourth paper are 100. Find the number of ways in which the candidate can score 60% marks in aggregate.

Solution

Let the marks scored by the candidate in four papers be x1, x2, x3 and x4

Then x1+ x2+x3 + x4 = 150 (60% of 250 = 150)

Where 0 £ x1, x2 , x3 £ 50 , 0 £ x4 £ 100.

The number of solutions of equation (1) is same as

The ceofficient of x150 in (1+ x + x2 +. . + x50)3 ( 1+ x + x2 +. . .+ x100).

= The ceofficient of x150 in

= The ceofficient of x150 in (1- x51)3 (1 – x101) (1 – x)-4

= The ceofficient of x150 in (1 – 3x51 + 3 x102 – x101) (1 – x)-4

= 153C3 – 3 102C3 + 3 51C3 – 52C3 = 110551.

 
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