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 Post subject: GMAT PermutationsPosted: Wed Oct 31, 2012 6:10 pm

Joined: Tue Nov 27, 2012 7:29 am
Posts: 12
Find the number of distinguishable permutations for the letters.
a) DOODLE
b) DECEMBER

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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.

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 Post subject: Re: GMAT PermutationsPosted: Wed Oct 31, 2012 6:37 pm

Joined: Sun May 30, 2010 2:23 am
Posts: 498
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
DDLEOO
DDLEOO

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

So 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.

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