In the below word how many words are there in which R and W are at the end positions?

RAINBOW

You need two points to draw a line. The order is not important. Line AB is the same as line BA. The problem is to select 2 points out of 3 to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines.

AB , AC

BA , BC

CA , CB

There is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB.

The lines are: AB, BC and AC ; 3 lines only.

So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important.

There are seven positions to be filled.

The first position can be filled using any of the 7 letters contained in PROBLEM.

The second position can be filled by the remaining 6 letters as the letters should not repeat.

The third position can be filled by the remaining 5 letters only and so on.

Therefore, the total number of ways of rearranging the 7 letter word = 7*6*5*4*3*2*1 = 7! ways.

ABACUS is a 6 letter word with 3 of the letters being vowels.

If the 3 vowels have to appear together, then there will 3 other consonants and a set of 3 vowels together.

These 4 elements can be rearranged in 4! Ways.

The 3 vowels can rearrange amongst themselves in 3!/2! ways as "a" appears twice.

Hence, the total number of rearrangements in which the vowels appear together are (4! x 3!)/2!

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.

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.