# GRE® Permutation Combination Practice

Concept: Reordering of n objects - all distinct. r! with constraints

This GRE quant practice question is a permutation combination problem solving question. An example of reordering n distinct elements with certain specified constraints.

#### Question: How many words can be formed by re-arranging the letters of the word PROBLEMS such that P and S occupy the first and last position respectively? (Note: The words thus formed need not be meaningful)

1. $\frac{8!}{2!}$
2. 8! - 2!
3. 6!
4. 8! - 2*7!
5. 6! * 2!

#### Video Explanation

Scroll down for explanatory answer text

#### Use these hints to get the answer

1. Concept: Number of ways r distinct objects can be reordered.
2. Make sure to keep the constraint that the first and last letter should be P and S respectively.
3. Find the number of possibilities with the constraint.

#### Reordering letters of PROBLEMS

'PROBLEMS' is an eight letter word where none of the letters repeat.

r objects, where all are distinct can be reordered in r! ways.

Therefore, if there had been no constraints, the letters of 'PROBLEMS' can be reordered in 8! ways.

However, the question states that 'P' and 'S' should occupy the first and last position respectively.

Therefore, the first and last position can be filled in only one way.

The remaining 6 positions in between P and S can be filled with the 6 letters in 6! ways.

#### Rearrangement Concepts : Number of ways to reorder objects

##### 1. Number of ways r distinct objects can be reordered

Let us say, three objects A, B, and C have to be reordered in 3 places 1, 2 and 3.

Any of the 3 letters can be placed in the 1st place.
If A occupies the 1st place, the 2nd place can be filled in only 2 ways - either B or C.
And the 3rd place can be filled in only 1 way - the only object that is yet to be placed.

Therefore, total number of ways = 3 ways to fill the 1st place AND 2 ways to fill the 2nd place AND 1 way to fill the 3rd.

So, total ways of reordering these 3 objects is 3 * 2 * 1 = 3! ways.

r distinct objects can be reordered in r! ways

The 6 different rearrangements of A, B, and C in the 3 places are ABC, ACB, BAC, BCA, CAB, and CBA.

##### 2. Number of ways r similar objects can be reordered

Let us say, we have to reorder A, A and A in 3 places 1, 2, and 3.

It is quite evident that any reordering that we try will look the same as AAA.

So, 3 similar objects can be reordered in only 1 way.

r similar objects can be reordered in only 1 way
##### 3. Number of ways r objects, of which x are alike can be reordered

Let us try and reorder A, B, and B in 3 places 1, 2, and 3

Possibilities 1st Place 2nd Place 3rd Place
1 A B B
2 B A B
3 B B A

There are only 3 possibilities instead of the 6 that would been possible had the 3 letters been different.

For A taking the 1st place, had the other two letters been B and C, these two letters could have reordered in places 2 and 3 in 2! ways. 2 distinct objects - 2! ways. But as the other two letters are B and B, they can reorder in only 1 way.

Therefore, we could count only 1 out of the 2! ways in which the two letters could have reordered had they been distinct.

So, the number of ways 3 objects of which 2 are alike can reorder can be expressed as $\frac{3!}{2!}$ ways.

r objects, of which x are alike, can be reordered in $\frac{r!}{x!}$ ways.