**Quote:**

Hi, I am in the GMAT prep course. I am learning permutations right now but got stuck on these two problems of permutations with repetitions. I don't know how the answers came out the way they did.

Let's solve the "a) DOODLE" problem first.

Imagine that all the letters in the word "DOODLE" are different. For example, let's color them.

DOODLE

Calculating the number of permutations of 6 different objects is simple:

6! = 6 × 5 × 4 × 3 × 2 × 1 = 720

The following "words" are among these permutations

DDLE

OODDLE

OOBut if the letters "D" were not colored, then this "words" would be just one word:

DDLE

OOSo each word with not colored D corresponds to 2 words with colored D. And instead of 6! variants we will get 6!/2 = 720/2 = 360 .

Now, if we decolour "O" as well, we will get 6!/(2 × 2) = 180 variants.

This solution should help you understand where the denominator in the formula n!/(s1! × s2! × ...) comes from.

Don't forget that we have factorials in the denominator.

In the second problem, "b) DECEMBER", the letter "E" is given three times. The rest of the letters appear only once. So the formula we use is

8! / 3! = (8 × 7 × 6 × 5 × 4 × 3 × 2 × 1) / (3 × 2 × 1) = 6720

One decolourized variant

EEEDCMBR

corresponds to 3! = 6 colorized variants (as if all the objects were different):

EEEDCMBR

EEEDCMBR

EEEDCMBR

EEEDCMBR

EEEDCMBR

EEEDCMBR

Remember, you do not need to colourize and decolourize objects each time, but understanding how the formula works will help you to remember it and use appropriately.