[vinvinvin]

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

(a) How many different possible arrangements are there?

14!/3!x2!x2! answers this question easy enough

(b) How many arrangements begin with P. The answer to the above /14 as there is only one P.

(c) How many arrangements start and end with O. As there are 3 O's the combinations of letters in between= 11!. This is is the right answer however this was kind of an educated guess so can anyone explain to me exactly why that was right.

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

To answer this Q I attempted the following. I imagined there was a space between each non-vowel and one before the beginning and end. Therefore 10 spaces.

To find the number of combinations of consonants I did the following,

9!/2!x3!

I then multiplied this by 10 x 9 x 8 x 7 x 6 as there is one less space each time a vowel fills a spot. I thought this would give me the final number of arrangements but this was not correct. Please can somebody advise me as to where I went wrong and how best to approach parts (c) and (d) of this question

[vinvinvin]

bump

[Azland]

For D

It might be simpler just to find out the number of possible arrangments when two vowels are next to each other and subtract it from your answer to 1

As for your educated guess, I dont know. That's as simple as it gets right? you explained the logic yourself.

[ian.slater]

(Original post by vinvinvin)

(c) How many arrangements start and end with O. As there are 3 O's the combinations of letters in between= 11!. This is is the right answer however this was kind of an educated guess so can anyone explain to me exactly why that was right.

I think there are 12 letters between the two O's. But the 12! includes double counting. So you need to deal with that. By coincidence it gives the same answer as you have.

[ghostwalker]

(Original post by vinvinvin)
(d) How many arrangements are there where no two vowels are next to each other?

To answer this Q I attempted the following. I imagined there was a space between each non-vowel and one before the beginning and end. Therefore 10 spaces.

To find the number of combinations of consonants I did the following,

9!/2!x3!

I then multiplied this by 10 x 9 x 8 x 7 x 6 as there is one less space each time a vowel fills a spot. I thought this would give me the final number of arrangements but this was not correct.

Your approach to (d) is along the right lines. The only thing you've forgotten is that there are 2 identical O's, hence divide by....

[TenOfThem]

(Original post by ghostwalker)
Your approach to (d) is along the right lines. The only thing you've forgotten is that there are 3 identical O's, hence divide by....

I can only see 2 Os in CONSTANTINOPLE

[TenOfThem]

(c)

There are 2 Os
These are at either end

there are 12 letters in between of which 2 are T and 3 are N

[ghostwalker]

(Original post by TenOfThem)
I can only see 2 Os in CONSTANTINOPLE

Clearly, I can't count.

[vinvinvin]

(Original post by ghostwalker)
Your approach to (d) is along the right lines. The only thing you've forgotten is that there are 2 identical O's, hence divide by....

Isn't 9!/2!x3! to find out the number of combinations of consonants though? Then as a vowel is added to a space it is 10 x 9 x 8 x 7 x 6 as the spaces decrease? What affect would the o's have if they are just filling the spaces?

[ghostwalker]

(Original post by vinvinvin)
Isn't 9!/2!x3! to find out the number of combinations of consonants though? Then as a vowel is added to a space it is 10 x 9 x 8 x 7 x 6 as the spaces decrease? What affect would the o's have if they are just filling the spaces?

If you had 5 different vowels, then your computation would be correct.

But having two O's, has the same effect as having two T's.