How many distinct permutations can be formed using the letters of the word prepared

Mathematics

1 answer:

Troyanec [42]2 years ago

5 0

Answer:

5,040

Step-by-step explanation:

Hmmm...this one is a bit tricky. First, prepared contains 8 letters. To find the number of ways this can be rearranged use the fundamental counting principle.

First letter - 8 choices

Second letter - 7 choices

Third letter - 6 choices

So the number of outcomes would be $8! = (8)(7)(6)(5)(4)(3)(2)(1)=40,320$

But wait, there's more. Prepared has some duplicate letters, the P, R, and E. That means if the letters were written ppeerrad it would be indistinguishable if it was the first or second P written first. So that means the letter P generates 2 outcomes that appear the same

$\frac{40,320}{2}=20,160$

Same for the letter R

$\frac{20,160}{2}=10,080$

And same for the letter E

$\frac{10,080}{2}=5,040$

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