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01. In how many different ways can the letters of the word 'MATHEMATICS' be arranged so that the vowels always come together? 
A. 10080 B. 4989600
C. 120960 D. None of these

Answer and Explanation

Answer: 120960

Explanation:
In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
Number of ways of arranging these letters = 8! / ((2!)(2!)) = 10080.

Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters = 4!/2! = 12.

Required number of words = (10080 x 12) = 120960.

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02. In how many different ways can the letters of the word 'DETAIL' be arranged in such a way that the vowels occupy only the odd positions?
A. 32 B. 48
C. 36 D. 60
E. 120

Answer and Explanation

Answer: 36

Explanation:
There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.
Let us mark these positions as under:
(1) (2) (3) (4) (5) (6)
Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5.
Number of ways of arranging the vowels = 3P3 = 3! = 6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = 3P3 = 3! = 6.
Total number of ways = (6 x 6) = 36.

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03. In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
A. 159 B. 194
C. 205 D. 209
E. None of these

Answer and Explanation

Answer: 209

Explanation:
We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
Required number
of ways = (6C1 x 4C3) + (6C2 x 4C2) + (6C3 x 4C1) + (6C4)
= (6C1 x 4C1) + (6C2 x 4C2) + (6C3 x 4C1) + (6C2)
= (6 x 4) + 6 x 5 x 4 x 3 + 6 x 5 x 4 x 4 + 6 x 5
2 x 1 2 x 1 3 x 2 x 1 2 x 1

= (24 + 90 + 80 + 15)
= 209.

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04. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done? 
A. 564 B. 645
C. 735 D. 756
E. None of these

Answer and Explanation

Answer: 756

Explanation:

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05. Find the number of ways in which a group of 4 boys and 3 girls can be selected out of a total of 9 boys and 9 girls.
A. 10584 B. 31824
C. 21168 D. None of these

Answer and Explanation

Answer: 10584

Explanation:
4 boys out of 9 can be selected in 9C4 ways = 126

3 girls out of 9 can be selected in 9C3 ways = 84

Required number of ways = 126*84 = 10584

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06. Find the number of ways in which 4 cattle can be selected from a group of 9 cows and 5 buffaloes, such that there is at least 1 buffalo present.
A. 900 B. 875
C. 1001 D. None of these

Answer and Explanation

Answer: 875

Explanation:
The minimum requirement is of 1 buffalo. Hence, 1 or more than 1 buffalo can be selected with no restriction on the number of cows. The possible combinations are:
(1 buffalo, 3 cows), (2 buffaloes, 2 cows), (3 buffaloes, 1 cow), (4 buffaloes, 0 cows).
Required number of ways of selection = (5C1*9C3) + (5C2*9C2) + (5C3*9C1) + (5C4)

= 420 + 360 + 90 + 5
= 875

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07. The number of ways of forming a committee of 6 members from a group of 4 men and 6 women is
A. 200 B. 210
C. 310 D. 220

Answer and Explanation

Answer: 210

Explanation:
The total number of members is 4 + 6 or 10. A committee of six
=>can be formed from 10 in 10C6 ways = (10C4) or 210 ways

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08. The number of rectangles that can be formed on a 8 × 8 chessboard is
A. 2194 B. 1284
C. 1196 D. 1296

Answer and Explanation

Answer: 1296

Explanation:
the number of rectangles that can be formed on an
=> n × n chessboard (R) = [n (n+1) / 2]2
=> As n = 8, R = 362 = 1296

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09. The number of distinct lines that can be formed by joining 20 points on a plane is of which no three points are collinear is
A. 190 B. 380
C. 360 D. 120

Answer and Explanation

Answer: 190

Explanation:
The number of lines that can be drawn, joining n points on a plane
 nC2 (when no three points are collinear). Here n = 20
required number of lines = 20C2 = 190

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10. Find the number of ways of studding 10 beads to form a necklace
A. 9! /2  B. 9! 
C. 9! 2! D. 10! / 2

Answer and Explanation

Answer: 9! /2 

Explanation:
n beads can be strung in a necklace in (n – 1)! / 2 ways
=> Here n = 10
Required number of ways = 9! /2

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