A

Selecting 2 boys from a group of 12 and correspondingly selecting 3 girls from group of 15 girls
[latex]^{12}{C}_{2} \times ^{15}{C}_{3} = 30030[/latex]
i.e, [latex]^{n}{C}_{r} = \frac{n!}{r! \times (n-r)!}[/latex]

C

Since the code has to be four lettered and need to have odd number of a’s, the code can have either 1 a or 3 a’s.
If the code has one a, then the remaining letters of the code are b, c and d.
∴ Number of different possible codes = 4! = 24
If the code has three a’s then remaining letter can be chosen in 3c1 = 3 ways
∴ Number of ways of forming the code = 3 × (4! ÷ 3!) = 12 (∵ 3 letters are repeated)
∴ Total number of ways of forming the code = 24 + 12 = 36 ways.

D

Every person shakes hand with every other person = 22 hand shakes
⇒ Total handshakes = 22 × 23
But A handshaking with B is same as B handshaking A
⇒ Total handshakes = [latex]\frac{(22 × 23)}{2}[/latex] = 253

D

Suppose we have A number of alphabets in a language.
⇒ Number of ways to make a 3 letter initial = A × A × A = [latex]{A}^{3}[/latex]
Given, number of ways to make 3 letter initial = 1 million = [latex]{10}^{6}[/latex]
⇒ [latex]{A}^{3}[/latex] = [latex]{10}^{6}[/latex]
⇒ A = [latex]{10}^{2}[/latex] = 100
∴ The language needs 100 alphabets

C

Number of ways 2 flavors can be selected out of 6 flavors = [latex]^{6}{C}_{2}[/latex] = 15
Number of ways pizza can be selected = 2 [there are 2 types of pizzas]
∴ Number of ways to select a pizza = 2 × 15 = 30 ways