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How many different four-letter words can be formed (the [#permalink]

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21 Feb 2005, 08:52

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How many different four-letter words can be formed (the words don't need to make sense) using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R?

A. 59
B. 11!/(2!*2!*2!)
C. 56
D. 23
E. 11!/(2!*2!*2!*3!)

We want four letter words that the first letter is E and the last letter is R.

So we need to get the two middle letters from the 11 letters (13-E and R): MDITERANEAN We have 8 different letters with E, A, N repeated twice. So it would be P(8,2)+3=56+3=59

(A)

Hong I followed the highlighted part , but then how come +3?????????
Could you plz explain that?

Last edited by Vijo on 25 Feb 2005, 03:54, edited 1 time in total.

Ok my thought is this. For the middle two letters, you could have two different ones from the eight different letters, or you could have two same ones from the three repeated letters. So the first part is C(8,2), and the second part is 3 (EE, AA, NN).

How many different four-letter words can be formed (the words don't need to make sense) using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R?

A. 59 B. 11!/(2!*2!*2!) C. 56 D. 23 E. 11!/(2!*2!*2!*3!)

how many four letter words if letters can be repeated?

How many different four-letter words can be formed (the words don't need to make sense) using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R?

A. 59 B. 11!/(2!*2!*2!) C. 56 D. 23 E. 11!/(2!*2!*2!*3!)

this is really a new concept in counting methods (permutation and combination), if hong's approach is the solution/answer.

How many different four-letter words can be formed (the words don't need to make sense) using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R?

A. 59 B. 11!/(2!*2!*2!) C. 56 D. 23 E. 11!/(2!*2!*2!*3!)

this is really a new concept in counting methods (permutation and combination), if hong's approach is the solution/answer.

why ? we have a four-letter-word that begins with a E and ends with a R => E _ _ R. The two spaces in the middle can be occupied by 8 distinct letters, hence 8P2. but its also possible to have repeated letters such as EE, AA and NN. thats why we add 3. i think its just slightly different to a classical permutation-letter-problem. _________________

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