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Find the number of distinct 6 letter words which the vowels appear next to each other in the word FERMAT

How do you do this if the letters have to be beside each other?

thanks in advance
 
anna said:
Find the number of distinct 6 letter words which the vowels appear next to each other in the word FERMAT

How do you do this if the letters have to be beside each other?

thanks in advance
Click to expand...

You just glue the E and the A together :D If you do that you get one new letter. It can be either AE or EA. You then have five letters (F, R, M, T and either AE or EA), so there are 5! possibilities for both AE and EA, so the total number of words you can make is 2*5!.
 
oh thats not hard then, so how bout if you had to keep letters seperate from one another, would that mean u add an extra letter or?
 
Assuming that you mean to find the number of ways in which the vowels are not together then think this way. You have just been given the number of ways in which the vowels are together. Thus if you subtract that number from the total rearrangements, 6!, would that be the number of ways that they are NOT together?
 
true Pka.

Umm same topic, I can't seem to get the right answer for this question here:
How many different 5-letter words can be formed using the letters of the word STEVIN, if both vowels must be used, they must be adjacent and E is to precede I?

Since EI have to be together, I counted them as one letter, and did 4!, but that's not right , GRR
 
What more problems can we think of involving good old FERMAT :D

What about this one: What is the number of 6 letters words you can make such that none of the letters are in the position they are in in the word FERMAT (i.e. every letter has to move relative to where it is in the word FERMAT).

Answer: 265.
 
anna said:
true Pka.

Umm same topic, I can't seem to get the right answer for this question here:
How many different 5-letter words can be formed using the letters of the word STEVIN, if both vowels must be used, they must be adjacent and E is to precede I?

Since EI have to be together, I counted them as one letter, and did 4!, but that's not right , GRR
Click to expand...

STEVIN has six letters, so you must leave one consonant out. You have four choices for that. The 4! you obtained must thus be multiplied by 4.
 
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