This question seems to be a bit typical, isn’t, but it is simplest.
1 red ball can be selected in 4C1 ways
1 white ball can be selected in 3C1 ways
1 blue ball can be selected in 2C1 ways

Total number of ways
= 4C1 x 3C1 x 2C1
= 4 x 3 x 2
= 24

Please note that we have multiplied the combination results, we use to add when there is OR condition, and we use to multiply when there is AND condition, In this question it is AND as
1 red AND 1 White AND 1 Blue, so we multiplied.

A number is divisible by 4 when its last two digits are divisible by 4
For this, the numbers should have their last two digits as 00, 04, 08, 12, 16, … 96

By the formula, a_n = a + (n-1)d
====>>>> 96 = 0 + (n-1)*4
==== >>>> n = 25

so there are 25 choices for last 2 digits and 9 choices (1-9) for the 1st digit
====>>>> so total 9*25 = 225

Here the order of choosing the elements doesn’t matter and this is a problem in combinations.

We have to find the number of ways of choosing 4 elements of this set which has 11 elements.

Maximum number of such different groups = ABC, ABD, ABE, BCE,BDE,CEA,DEA =7.

The man needs to take 6 steps to cross the river. He can do this in the following ways:

===>>> Crossing the river by 6 unit steps = 1 way.

===>>> Crossing the river by 4 unit steps and 1 double step = ⁵C₁ = 5C4 = 5 ways.

===>>> Crossing the river by 2 unit steps and 2 double steps = ⁴C₂ = 6 ways.

===>>> Crossing the river by 3 double steps = 1 way.

Hence, the required number of ways = 1 + 5 + 6 + 1 = 13.

1 million distinct 3 digit initials are needed.
Let the number of required alphabets in the language be ‘n’.
=> Therefore, using ‘n’ alphabets we can form n * n * n = n³ distinct 3 digit initials.
=> Note distinct initials is different from initials where the digits are different.
=> For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different.
=>This n³ different initials = 1 million
======>>>>> i.e. n³ = 10⁶ (1 million = 10⁶)
======>>>>> n³ = (10²)³ => n = 10² = 100
Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.

Women can select 3 chairs from chairs numbered 1 to 5 in ⁵C₃ ways and remaining 6 chairs can be selected by 4 men in ⁶C₄ ways. Hence the required number of ways = ⁵C₃ × ⁶C₄

The first two places can only be filled by 3 and 5 respectively and there is only 1 way for doing this.

Given that no digit appears more than once. Hence we have 8 digits remaining $(0,1,2,4,6,7,8,9)$

So, the next 4 places can be filled with the remaining 8 digits in ⁸P₄ ways.

Total number of ways = ⁸P₄ $=8\times 7\times 6\times 5=1680$

5!*3! = 120*6 = 720

The first letter is E and the last one is R.

Therefore, one has to find two more letters from the remaining 11 letters. Of the 11 letters, there are 2 Ns, 2Es and 2As and one each of the remaining 5 letters.

The second and third positions can either have two different letters or have both the letters to be the same.

Case 1: When the two letters are different. One has to choose two different letters from the 8 available different choices. This can be done in 8 * 7 = 56 ways.

Case 2: When the two letters are same. There are 3 options – the three can be either Ns or Es or As. Therefore, 3 ways.

Total number of possibilities = 56 + 3 = 59

The group contained six men and seven women

Three men can be selected from six men in ⁶C₃ ways.

Four women can be selected from seven women in ⁷C₄ ways.

There are three cases:

1. 3 green coins
2. 2 green coins + 1 non-green coin
3. 1 green coin + 2 non-green coins

Therefore, total number of ways = ³C₃ + ³C₂ * ⁶C₁ + ³C₁ * ⁶C₂ = 1 + 3*6 + 3*15 = 64.