If a card hand that consists of four Queens and an Ace is rearranged, nothing has changed.

The hand still contains four Queens and an Ace. Thus, use the combination formula for problems with cards.

We have 4 eights and 4 sevens.

We want 3 eights and 2 sevens.

C(have 4 eights, want 3 eights) x C(have 4 sevens, want 2 sevens) 

C(4,3) x C(4,2) = 24

Therefore there are 24 different ways in which to deal the desired hand.

Since order does not matter it is a combination. 

The word AND means multiply. 

Given 4 basketball, 3 volleyball, 2 soccer. 

We want 2 basketball games and 1 other event. There are 5 choices left. 

C(n,r) 

C(How many do you have, How many do you want) 

C(have 4 basketball, want 2 basketball) x C(have 5 choices left, want 1) 

C(4,2) x C(5,1) = (6)(5) = 30

Therefore there are 30 different ways in which you can go to two basketball games and one of the other events.

Since it does not matter what order the committee members are chosen in, the combination formula is used.

Committees are always a combination unless the problem states that someone like a president has higher hierarchy over another person. If the committee is ordered, then it is a permutation.

C(17,7)= 19,448

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)

  => n = 102 = 100

Hence, the language needs to have a minimum of 100 alphabets to achieve the objective.

Required number of ways = (7C5*3C2) = (7C2*3C1) = 63

When 4 dice are rolled simultaneously, there will be a total of 6 x 6 x 6 x 6 = 1296 outcomes.

The number of outcomes in which none of the 4 dice show 3 will be 5 x 5 x 5 x 5 = 625 outcomes.

Therefore, the number of outcomes in which at least one die will show 3 = 1296 – 625 = 671

From 5 consonants, 3 consonants can be selected in 5C3 ways.

From 4 vowels, 2 vowels can be selected in 4C2 ways.

Now with every selection, number of ways of arranging 5 letters is 5P5ways.

Total number of words = 5C3*4C2*5P5

                                = 10x 6 x 5 x 4 x 3 x 2 x 1= 7200

4 novels can be selected out of 5 in 5C4 ways.

2 biographies can be selected out of 4 in 4C2 ways.

Number of ways of arranging novels and biographies = 5C4*4C2  = 30

After selecting any 6 books (4 novels and 2 biographies) in one of the 30 ways, they can be arranged on the shelf in 6! = 720 ways.

By the Counting Principle, the total number of arrangements = 30 x 720 = 21600