# Permutation doubt

• Jun 5th 2009, 11:38 PM
mj.alawami
Permutation doubt
Question
How many permutations can be made from the letters in the word "Wednesday" if a)4 letters are used at the same time b) all the letters are used

Attempt
a) $\frac{9!}{(9-4)!} = 3024$

b) Since some letters are alike
$\frac{9!}{1!*2!*2!*1!*1!*1!*1!} =90720$

Am I correct?
• Jun 6th 2009, 03:54 AM
Plato
Quote:

Originally Posted by mj.alawami
Question
How many permutations can be made from the letters in the word "Wednesday" if a)4 letters are used at the same time b) all the letters are used Attempt
a) $\frac{9!}{(9-4)!} = 3024$

b) Since some letters are alike
$\frac{9!}{1!*2!*2!*1!*1!*1!*1!} =90720$

The answer for part (b) is correct.

For part (a), there are several cases to consider.
i) all letters are different.
ii) use either 'e' or 'd' twice, the others are different.
iii) have two twice 'eded'
• Jun 6th 2009, 06:56 AM
mj.alawami
Quote:

Originally Posted by Plato
The answer for part (b) is correct.

For part (a), there are several cases to consider.
i) all letters are different.
ii) use either 'e' or 'd' twice, the others are different.
iii) have two twice 'eded'

So how would the answer of part A come out ... (Wondering)

Because I am getting confused or what you are trying to say is that the question is wrong?

• Jun 6th 2009, 07:07 AM
Plato
For case i) $P(7,4)=\frac{7!}{3!}$

For case ii) $2 \binom{5}{2} \left(\frac{4!}{2}\right)$

For case iii) $\frac{4!}{(2!)^2}$