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

There is 1 way to correctly guess who comes in first, second, and third. There is only one set of first, second and third place winners. You must correctly guess these three people, and there is only one way to do so.

There are 5 slots.

                   __ __ __ __ __

The first slot must be a four. There are 4 ways to put a four in the first slot.

There are 4 ways to put a five in the second slot, and there are 4 ways to put a six in the third slot. etc.

(4)(4)(4)(4)(4) = 1024

Therefore there are 1024 different ways to produce the desired hand of cards.

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.

Out of 26 alphabets two distinct letters can be chosen in 26P2 ways. Coming to numbers part, there are 10 ways.(any number from 0 to 9 can be chosen) to choose the first digit and similarly another 10ways to choose the second digit. Hence there are totally 10X10 = 100 ways. 

Combined with letters there are 6P2 X 100 ways = 65000 ways to choose vehicle numbers.

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.

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

FV1
= Future value of $2000, 1 year later

= PV (1+  i)^n