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?

1. 3! 4! 8! 4!
2. 3! 8!
3. 4! 4!
4. 8! 4! 4!

Answer:Option 1

Explanation:

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.

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

1. 600
2. 120
3. 240
4. 60

Answer:Option 1

Explanation:

Solution:

APPLE = 5 letters.
But two letters PP is of same kind.
Thus, required permutations,
= 5!/2!
= 120/2
= 60.

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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.

1. 7560,60,1680
2. 7890,120,650
3. 7650,200,4444
4. None of these

Answer:Option 1

Explanation:

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.

(a) There are 4 vowels and all are alike i.e. 4A's.
_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.

(b) Taking both L's together and treating them as one letter we have 8 letters out of which A repeats 4 times and others are distinct.
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.

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

1. 144
2. 288
3. 12
4. 256

Answer:Option 1

Explanation:

Solution:

Let the Arrangement be,
B G B G B G B
4 boys can be seated in 4! Ways.
Girl can be seated in 3! Ways.
Required number of ways,
= 4!*3! = 144.

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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 ?

1. 11
2. 12
3. 13
4. 14

Answer:Option 1

Explanation:

Solution:

⁵C₁ * ³C₁ -1
= 15-1
= 14

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

1. 190
2. 180
3. 240
4. 390

Answer:Option 1

Explanation:

Solution:

Number of ways = ²⁰C₂ = 20!/(2! 18!)
= 20*19/2
= 190.

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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?

1. 9
2. 30
3. 36
4. 15

Answer:Option 1

Explanation:

Solution:

There are 6 candidates and a voter has to vote for any two of them.
So, the required number of ways is,
= ⁶C₂ = 6!/2!*4!
= 15.

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

1. 56
2. 24
3. 16
4. 8

Answer:Option 1

Explanation:

Solution:

A triangle needs 3 points.
And polygon of 8 sides has 8 angular points.
Hence, number of triangle formed,
= ⁸C₃ = (8*7*6)/(1*2*3)
= 56.

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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.

1. 120
2. 116
3. 140
4. 20

Answer:Option 1

Explanation:

Solution:

The number of triangle can be formed by 10 points = ¹⁰C₃.
Similarly, the number of triangle can be formed by 4 points when no one is collinear=⁴C₃.
In the question, given 4 points are collinear, Thus, required number of triangle can be formed,
= ¹⁰C₃-⁴C₃ = 120-4 = 116.

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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.

1. 15
2. 14
3. 21
4. 25

Answer:Option 1

Explanation:

Solution:

Let n be the number of persons in the party. Number of hands shake = 105; Total number of hands shake is given by ⁿC₂.

Now, according to the question,

ⁿC₂ = 105;
Or, n!/[2!*(n-2)!] = 105;
Or, n*(n-1)/2 = 105;
Or, n²-n = 210;
Or, n²-n-210 = 0;
Or, n = 15, -14;

But, we cannot take negative value of n.
So, n = 15 i.e. number of persons in the party = 15.

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