S1 Permutations Question

Hi,
I'm stuck on a challenging permutations question, I have the answer but I'm not quite getting there, here goes:

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.

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

ANSWER: 457,228,800

Reply 1

TSR Jessica

19

Sorry you've not had any responses about this. :frown:

Are you sure you’ve posted in the right place? Posting in the specific Study Help forum should help get responses. :redface:

I'm going to quote in Puddles the Monkey now so she can move your thread to the right place if it's needed. :h:

Spoiler

Reply 2

Karlw1lk1ns

OP

1

Thanks

... the question is still pestering me!

Reply 3

M.C. Math

7

Original post by TSR Jessica

Sorry you've not had any responses about this. :frown:

I haven't cracked this one. Will keep trying, but am looking forward to see how it's done.

Reply 4

Karlw1lk1ns

OP

1

Original post by M.C. Math

I haven't cracked this one. Will keep trying, but am looking forward to see how it's done.

Me too, its bugging me will keep trying.

Reply 5

M.C. Math

7

Original post by Karlw1lk1ns

Me too, its bugging me will keep trying.

Hi. I think I've come up with a solution...

The vowels have to be at least one character apart, so the first few positions are
1st, 3rd, 5th, 7th, 9th
1st, 3rd, 5th, 7th, 10th
1st, 3rd, 5th, 7th, 11th
1st, 3rd, 5th, 7th, 12th
1st, 3rd, 5th, 7th, 13th
1st, 3rd, 5th, 7th, 14th
2nd, 4th, 6th, 8th, 10th
2nd, 4th, 6th, 8th, 11th
etc
up to
1st, 8th, 10th, 12th, 14th
2nd, 8th, 10th, 12th, 14th
3rd, 8th, 10th, 12th, 14th
4th, 8th, 10th, 12th, 14th
5th, 8th, 10th, 12th, 14th
6th, 8th, 10th, 12th, 14th
The total number of combinations is 10C5, i.e. 252 (10 positions of which 5 are chosen, order not important)

For each position, there are 5! arrangements of the 5 vowels, but O is repeated twice, so there are 5!/2! arrangements, i.e. 60. So now we have a total number of arrangements of
10C5 x (5!/2!), 252 x 60 to give 15120 arrangements of the vowels.

Now to the consonants. There are 9 of them, so 9! arrangements. However N appears three times and T appears twice, so there are 9!/(3! x 2!) arrangements, i.e. 30240.

Putting the vowels and consonants together gives a total number of arrangements of
252 x 60 x 30240 = 457228800.

There may be an alternative/better way of doing it so I'm open to suggestions! :smile:

Reply 6

Karlw1lk1ns

OP

1

Original post by M.C. Math

Hi. I think I've come up with a solution...

The vowels have to be at least one character apart, so the first few positions are
1st, 3rd, 5th, 7th, 9th
1st, 3rd, 5th, 7th, 10th
1st, 3rd, 5th, 7th, 11th
1st, 3rd, 5th, 7th, 12th
1st, 3rd, 5th, 7th, 13th
1st, 3rd, 5th, 7th, 14th
2nd, 4th, 6th, 8th, 10th
2nd, 4th, 6th, 8th, 11th
etc
up to
1st, 8th, 10th, 12th, 14th
2nd, 8th, 10th, 12th, 14th
3rd, 8th, 10th, 12th, 14th
4th, 8th, 10th, 12th, 14th
5th, 8th, 10th, 12th, 14th
6th, 8th, 10th, 12th, 14th
The total number of combinations is 10C5, i.e. 252 (10 positions of which 5 are chosen, order not important)

For each position, there are 5! arrangements of the 5 vowels, but O is repeated twice, so there are 5!/2! arrangements, i.e. 60. So now we have a total number of arrangements of
10C5 x (5!/2!), 252 x 60 to give 15120 arrangements of the vowels.

Now to the consonants. There are 9 of them, so 9! arrangements. However N appears three times and T appears twice, so there are 9!/(3! x 2!) arrangements, i.e. 30240.

Putting the vowels and consonants together gives a total number of arrangements of
252 x 60 x 30240 = 457228800.

There may be an alternative/better way of doing it so I'm open to suggestions! :smile:

You're a god ! :smile:

Thanks this has been bugging me. It was a starred question in the textbook so I think there are no quicker ways of doing it

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