[SOLVED] Permutations no two objects together arrangment..

The letter of the word CONSTANTINOPLE are written on 14cards, one on each card. the cards are shuffled and then arranged in a straight line.

(d) How many arrangements are there when no two vowels are next to each other?

There are 5 vowels here as I see it, A, I, E, 2 x O.

If we let vowels = V and consonants = C, then an arrangement like this can happen:

We have (14 - 5) = 9 consonants

- C1 - C2 - C3 - C4 - C5 - C6 - C7 - C8 - C9 -

Now, we have 10 places to input the first vowel, 9 for 2nd, 8 for 3rd, 7 for 4th, 6 for 5th ...

I am not sure how to solve it next. I did (9! / (3! 2!)) x 10 x 9 x 8 x 7 x 6 but that doesn't work and I am not sure what am I doing wrong. (Headbang)

Arrangements with repeated letters

Hello struck Quote:

Originally Posted by

**struck** The letter of the word CONSTANTINOPLE are written on 14cards, one on each card. the cards are shuffled and then arranged in a straight line.

(d) How many arrangements are there when no two vowels are next to each other?

There are 5 vowels here as I see it, A, I, E, 2 x O.

If we let vowels = V and consonants = C, then an arrangement like this can happen:

We have (14 - 5) = 9 consonants

- C1 - C2 - C3 - C4 - C5 - C6 - C7 - C8 - C9 -

Now, we have 10 places to input the first vowel, 9 for 2nd, 8 for 3rd, 7 for 4th, 6 for 5th ...

I am not sure how to solve it next. I did (9! / (3! 2!)) x 10 x 9 x 8 x 7 x 6 but that doesn't work and I am not sure what am I doing wrong. (Headbang)

Your reasoning is OK as far as you've gone. The consonants, as you say, can be arranged in $\displaystyle \frac{9!}{3!2!}$ ways, and there are then 10 slots for the vowels. So we next choose 5 slots from these 10 in $\displaystyle ^{10}C_5$ ways, and then place the vowels in them in $\displaystyle \frac{5!}{2!}$ ways.

Total: $\displaystyle \frac{9!}{3!2!}\times \frac{10!}{5!5!} \times \frac{5!}{2!} = \frac{9!10!}{3!2!5!2!} = 457,228,800$ ways.

How does that seem?

Grandad