Question

How many different four letter words can be formed (the words need not be meaningful) using the letters of the word PACIFIC such that the first letter is P and the last letter is F?

Answer 1

Answer:

If repetition is allowed

Letters available to fill the space are P, A, C, I and F. (Doppelgangers of C and I don't matter)

Total number of arrangements = 5×5=25

If repetition is NOT allowed

Letters available for arrangement are A,C,C,I and I. (P and F are already used)

First, let's consider only A, C and I.

Number of arrangements = 3×2=6

And there are other 2 cases are where letters 'C' and 'I' repeat twice in each respective case.

So, total number of arrangements = 8

Answer 2

Given:

The given word = PACIFIC.

To Find:

We have to find out the number of four-letter words that can be formed using the letters of the word PACIFIC such that the first letter is P and the last letter is F.

Solution:

The number of letters in the word PACIFIC = 7.

With P as the first letter and F as the last letter, the remaining letters are A, C, I, I, and C.

We have to make four-letter words.

That is, we have to arrange these 5 letters in two places to form a four-letter word that begins with "P" and ends with "F".

The number of ways in which these 5 letters can be arranged between P and F = $^{n}P_r.$.

Substituting the values of n and r in the above equation as, n = 5 and r = 2, we get,

$^{n}P_r = ^{5}P_2 = \frac{5!}{(5-2)!} = \frac{120}{6} = 20.$

Hence, the number of four-letter words that can be formed using the letters of the word PACIFIC such that the first letter is P and the last letter is F is  20.

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