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Solved Subjective Question on Permutations and Combinations Set 4

Posted on - 04-01-2017

Math

IIT JEE

Example.1

Let A = {1, 2, 3, .... ,n}. If ai is the minimum element of the set Ai (where Ai denotes the subset of A containing exactly three elements) and X denotes the set of Ai s, then evaluate .

Solution

Since we can take three elements out of n in nC3 ways, X will contains nC3 subsets. There are exactly n-1C2 elements of X having 1 as least element; exactly n-2C2 elements of X having 2 as least element and so on. In general there are exactly n-rC2 elements of X having r
(1 £ r £ n – 2) as least element
. Thus,.

= 1(n-1C2)+2(n-2C2) + 3(n-3C2) + …. + (n –2) (2C2)

=

=

=

=

=

=

=

= .

Example.2

There are 15 seats in a row numbered as 1 to 15. In how many ways 4 persons can sit such that seat number 6 is always occupied in no two persons sit in adjacent seats.

Solution

Since seat number ‘6’ is always occupied remaining 3 can sit in the following ways Case-I :

2 left to 6 and 1 right to 6. .

Two can sit in the ways .

(1, 3) or (1, 4) or (2, 4) and one who is sitting right to 6 can sit in the 8 ways. Hence Total number of ways in this case = 3 ´ 8 = 24.

Case–II :

1 left to 6 and 2 right to 6

The person sitting left can sit in 4 ways

Number of ways in which two person can sit right to 6 is same as number of integral solution of x1 + x2 + x3 = 7

x1 ≥ 1, x2 ≥ 1, x3 ≥ 0

i.e coefficient of t5 in (1 –t)–3 = 5 +3 –1C3–1 = 7C2 = 21.

Hence total number of ways =21´4 = 84 ways.

Case–III :

No one sits left to 6 and 3 sit right to 6. .

Number of ways.

= Number of solution to x1 +x2 +x3 +x4 = 6(x1 ≥1, x2 >1, x3 ≥ 1, x4≥ 1)

= coefficient of t3 in (1 –t)-4 = 4+3-1C4-1 = 6C3 = 20

Hence total number of ways in this case = 20

Hence number of ways of selection of seats = 24 +84 +20 =128

Total number of ways of arrangement of 4 person on these seats

= 128 ´ 4! = 3072

Example.3

There are n straight lines in a plane, no two of which are parallel and no three of which pass through the same point. How many additional lines can be generated by means of points of intersections of the given lines. .

Solution

Let AB be any one of the n straight lines and let it be intersected by some other line CD at point E.

Note that line AB contains (n – 1) such points. It follows that there are n (n – 1) such points of intersection. Since each of the points of intersection occurs in two of the n lines, the number of such points is . We shall now find the number of additional lines passing through these points. The number of additional lines passing through points of the type E is equal to the number of points lying outside the line AB and CD, because a new line is obtained through E only if the other point lies outside AB and CD. Since AB and CD each contain
(n – 2) new points, the number of new points outside AB and CD is

Thus, the number of lines through E is . Since there are new points, the number of new lines is

But in this way each line is counted twice. Therefore, the required number of lines is

=

Example.4

(a) In how many ways can 2n people be divided into n pairs?

(b) In how many ways can you equally distribute 100 packages of food to 10 refugees?

(c) In how many ways mn things be equally distributed in n groups ?

(d) Find the number of ways of selecting r pairs out of n different things.

(e) In how many ways can 22 books be divided into 5 students so that two students will have 5 books each and the other three students will have 4 books each.

(f) (i) In how many ways can a pack of 52 cards be divided equally among four players in order?

(ii) In how many ways can you divide these cards into 4 sets, three of them having 17 cards each and the fourth just one card?

Solution

(a) No. of permutations of 2n people (2n)! Now the two people belonging to every pair can be considered as alike and also the n pairs of same size can be considered alike i.e. there is no arrangement of groups.

\ No. of ways =

(b) No of permutations of 100 packages = 100! The 10 packages given to each refugee can be considered as alibi. No. of ways of making 10 groups consisting of 10 packages each =

Now distribution of packages can be done in

(c) mn thing can be arranged in (mn)! ways. Of these, n groups of m things each can be treated as identical and also the m thing belong to each group can be treated as identical.

So, number of ways = .

(d) To make r pairs, first select 2r objects out of n and then divided them into pairs.

Required number of ways = nC2r .

(e) No. of permutations of 22 books = 22! of these, two groups of 5 books each and 3 groups of 4 books each can be treated as identical.

\ no. of ways of division into groups =
Now these groups have to be distributed among 5 students

Required number of ways = .

(f) (i) For the first player we have 52C13 choices, for the second player 39C13 choices, for the third player 26C13 choices and for the last player we have 13C13 choices. Hence, the total number of choices.

= (52C13 ) (39C13 ) (26C13 ) (13C13 )

= .

(ii) For the first set we have 52C17 choices, for the second set 35C17 choices, for the third set 18C17 choices and for the last set we have 1C1 choices. But the first three sets can be interchanged in 3! ways. Hence, the total number of ways for the required distribution.

= (52C17 ) (35C17 ) (18C17 ) (1C1 )

Example.5

A test has 4 parts The first 3 parts carry 10 marks each and the 4th part carries 20 marks. Assuming that marks are not given in fractions, find the number of ways in which a candidate can get 30 marks out of 50.

Solution

Part

Max. Marks.

Marks Secured

1

10

x1

2

10

x2

3

10

x3

4

20

x4

Where 0 £x1 £ 10, 0 £ x2 £ 10

0 £ x3 £ 10, 0 £ x4 £ 20

\ No. of ways in which the candidate can get 30 marks.

= Coefficient of x30 in the expression

(x° + x1 + x2 + …. + x10)3 (x° + x1 + x2 + …. + x20).

= Coefficient of x30 in

= Coefficient of x30 in (1 – x11)3 (1 – x21) (1 – x)-4

= Coefficient of x30 in (1 – 3x11 + 3x22) (1 – x21) (1 – x)-4

= Coefficient of x30 in (1 – 3x11 + 3x22 – x21) (1 – x)-4

= + (-3)+ (3)+ (-1)

= 1111.

 
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