Answer: Option B

Explanation:
From a group of 7 men and 6 women, five persons are to be selected with at least 3 men.
Hence we have the following 3 choices
We can select 5 men ------(Option 1)
Number of ways to do this = 7_C5
We can select 4 men and 1 woman ------(Option 2) 
Number of ways to do this = 7_C4 x 6_C1
We can select 3 men and 2 women ------(Option 3) 
Number of ways to do this = 7_C3 x 6_C2 
Total number of ways 
= 7C5 + [7_C4 x 6C1] + [7_C3 x 6_C2]
= 7_C2 + [7_C3 x 6_C1] + [7_C3 x 6_C2] [Applied the formula nCr = n_{C(n - r)} ]
=[(7×6)/(2×1)]+[(7×6×5)/(3×2×1)×6]+[(7×6×5)/(3×2×1)×(6×5)/(2×1)]
= 21 + 210 + 525 = 756

Answer: Option B

Explanation:
The word 'OPTICAL' has 7 letters. It has the vowels 'O','I','A' in it and these 3 vowels 
should always come together. Hence these three vowels can be grouped and considered as a
single letter. That is, PTCL(OIA). 
Hence we can assume total letters as 5. and all these letters are different.
Number of ways to arrange these letters = 5! = [5 x 4 x 3 x 2 x 1] = 120
All The 3 vowels (OIA) are different
Number of ways to arrange these vowels among themselves = 3! = [3 x 2 x 1] = 6
Hence, required number of ways = 120 x 6 = 720

Answer: Option A

Explanation:

The word 'CORPORATION' has 11 letters. It has the vowels 'O','O','A','I','O' in it and 
these 5 vowels should always come together. Hence these 5 vowels can be grouped 
and considered as a single letter. that is, CRPRTN(OOAIO). 
Hence we can assume total letters as 7. But in these 7 letters, 'R' occurs 2 times and
rest of the letters are different.
Number of ways to arrange these letters = 7!2! = (7×6×5×4×3×2×1)/(2×1) = 2520
In the 5 vowels (OOAIO), 'O' occurs 3 and rest of the vowels are different.
Number of ways to arrange these vowels among themselves = 5!3! = (5×4×3×2×1)/(3×2×1) = 20
Hence, required number of ways = 2520 x 20 = 50400

Answer: Option A

Explanation:
In a circle, 5 boys can be arranged in 4! ways
Given that the boys and the girls alternate. 
Hence there are 5 places for girls which can be arranged in 5! ways
Total number of ways = 4! x 5! = 24 x 120 = 2880

Answer: Option C

Explanation:
Total number of ways 
= 10_C8 x 10_C4
= 10_C2 x 10_C4 [Applied the formula nCr = n_{C(n - r)} ]
=(10×9)/(2×1)(10×9×8×7)/(4×3×2×1)
=45 x 210
= 9450

Answer: Option A

Explanation:
1 red ball can be selected in 4_C1 ways
1 white ball can be selected in 3_C1 ways
1 blue ball can be selected in 2_C1 ways
Total number of ways 
= 4_C1 x 3_C1 x 2_C1
=4 x 3 x 2
= 24

Answer: Option B

Explanation:
He can go in any car out of the 25 cars. 
Hence He can go in 25 ways.
Since he can not come back in the same car that he used for travelling,
He can return in 24 ways.
Total number of ways = 25 x 24 = 600

Answer: Option B

Explanation:
He has has ten patterns of chairs and eight patterns of tables
Hence, A chair can be arranged in 10 ways and 
A table can be arranged in 8 ways
Hence one chair and one table can be arranged in 10 x 8 ways = 80 ways

Answer: Option B

Explanation:
5 subjects can be arranged in 6 periods in 6_P5 ways.
Remaining 1 period can be arranged in 5_P1 ways.
Two subjects are alike in each of the arrangement. So we need to divide by 2! to avoid overcounting.
Total number of arrangements = (6_P5 x 5_P1)/2! = 1800
Alternatively this can be derived using the following approach.
5 subjects can be selected in 5_C5 ways.
Remaining 1 subject can be selected in 5C1 ways.
These 6 subjects can be arranged themselves in 6! ways.
Since two subjects are same, we need to divide by 2!
Total number of arrangements = (5C5 × 5C1 × 6!)/2! = 1800

Answer: Option A

Explanation:
100_P2 = 100 x 99 = 9900