# permutation - basic

• July 23rd 2010, 10:19 PM
nikk
permutation - basic
below the basic question of permutation:

given that 4 letters are chosen from the word PAYMENT,find the number of arrangement that start with a vowel

ans: 2P1 x 6P3 = 240
my ans: 120

any guide?(Punch)
• July 23rd 2010, 10:31 PM
pickslides
It seems like the question assumes the only vowel can be in the first postition.

Giving $^2P_1 \times ^6P_3 = 240$

As you may realise you start the 4 letter arrangement with a vowel and have another vowel (as there are 2) in positions 2-4.
• July 23rd 2010, 10:38 PM
nikk
tq for your post, but i hope u can commnet my ans attached as below:

http://img839.imageshack.us/img839/7452/99010889.jpg
• July 23rd 2010, 10:50 PM
pickslides
Why can't N or T be in the last postition? I think you are calculating combinations not permutations as you should. Are you aware of the difference?
• July 23rd 2010, 10:54 PM
nikk
it mean what i am doing is combination?
• July 24th 2010, 06:49 AM
Soroban
Hello, nikk!

Why do they include $Y$ in these problems?
There is always that controversy: "Is $Y$ a vowel?"

Quote:

Given that 4 letters are chosen from the word PAYMENT,
find the number of arrangement that start with a vowel.

If $Y$ is a vowel: . $\;\underbrace{\text{vowel}}_3 \text{ - } \underbrace{\text{any}}_6 \text{ - } \underbrace{\text{any}}_5 \text{ - }\underbrace{\text{any}}_4 \;=\;360$

If $Y$ is not a vowel: . $\;\underbrace{\text{vowel}}_2 \text{ - } \underbrace{\text{any}}_6 \text{ - } \underbrace{\text{any}}_5 \text{ - }\underbrace{\text{any}}_4 \;=\;240$

• July 25th 2010, 07:54 AM
nikk
tq ror your idea..nice