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Permutation and Combination

1.In how many ways can 8 Indians and, 4 American and 4 Englishmen can be seated in a row so that all person of the same nationality sit together?

**Answer:**Option** 1**

Solution:

Taking all person of same nationality as one person, then we will have only three people.These three person can be arranged themselves in 3! Ways.

8 Indians can be arranged themselves in 8! Way.

4 American can be arranged themselves in 4! Ways.

4 Englishman can be arranged themselves in 4! Ways.

Hence, required number of ways = 3!*8!*4!*4! Ways.

2.How many Permutations of the letters of the word APPLE are there?

**Answer:**Option** 1**

Solution:

APPLE = 5 letters. But two letters PP is of same kind.

Thus, required permutations,

= 5!/2!

= 120/2

= 60.

3.How many different words can be formed using all the letters of the word ALLAHABAD?

(a) When vowels occupy the even positions.

(b) Both L do not occur together.

(a) When vowels occupy the even positions.

(b) Both L do not occur together.

**Answer:**Option** 1**

Solution:

ALLAHABAD = 9 letters. Out of these 9 letters there is 4 A's and 2 L's are there.So, permutations = 9!/4!.2!= 7560.

_2nd _4th _6th _8th _

These even places can be occupied by 4 vowels. In 4!/4! = 1 Way.

In other five places 5 other letter can be occupied of which two are alike i.e. 2L's.

Number of ways = 5!/2! Ways.

Hence, total number of ways in which vowels occupy the even places = 5!/2! *1 = 60 ways.

These 8 letters can be arranged in 8!/4! = 1680 ways.

Also two L can be arranged themselves in 2! ways.

So, Total no. of ways in which L are together = 1680 * 2 = 3360 ways.

Now,

Total arrangement in which L never occur together,

= Total arrangement - Total no. of ways in which L occur together.

= 7560 - 3360 = 4200 ways.

4.In how many ways 4 boys and 3 girls can be seated in a row so that they are alternate.

**Answer:**Option** 1**

Solution:

Let the Arrangement be,4 boys can be seated in 4! Ways.

Girl can be seated in 3! Ways.

Required number of ways,

= 4!*3! = 144.

5.A two member committee comprising of one male and one female member is to be constitute out of five males and three females. Amongst the females. Ms. A refuses to be a member of the committee in which Mr. B is taken as the member. In how many different ways can the committee be constituted ?

**Answer:**Option** 1**

Solution:

= 15-1

= 14

6.In how many ways 2 students can be chosen from the class of 20 students?

**Answer:**Option** 1**

Solution:

Number of ways = = 20*19/2

= 190.

7.Three gentlemen and three ladies are candidates for two vacancies. A voter has to vote for two candidates. In how many ways can one cast his vote?

**Answer:**Option** 1**

Solution:

There are 6 candidates and a voter has to vote for any two of them.So, the required number of ways is,

=

= 15.

8.Find the number of triangles which can be formed by joining the angular points of a polygon of 8 sides as vertices.

**Answer:**Option** 1**

Solution:

A triangle needs 3 points.And polygon of 8 sides has 8 angular points.

Hence, number of triangle formed,

=

= 56.

9.There are 10 points in a plane out of which 4 are collinear. Find the number of triangles formed by the points as vertices.

**Answer:**Option** 1**

Solution:

The number of triangle can be formed by 10 points = Similarly, the number of triangle can be formed by 4 points when no one is collinear=

In the question, given 4 points are collinear, Thus, required number of triangle can be formed,

=

10. In a party every person shakes hands with every other person. If there are 105 hands shakes, find the number of person in the party.

**Answer:**Option** 1**

Solution:

Let n be the number of persons in the party.
Number of hands shake = 105;
Total number of hands shake is given by Now, according to the question,

Or, n!/[2!*(n-2)!] = 105;

Or, n*(n-1)/2 = 105;

Or, n

Or, n

Or, n = 15, -14;

But, we cannot take negative value of n.

So, n = 15 i.e. number of persons in the party = 15.