Question

How many distinct 9-letter words starting with “i” can be formed from the word “committee”?

Answer 1

We have already chosen "i" to be the first letter. This means that we still have 8 letters that we can use as the second letter. Once that letter is chosen, we still have 7 letters that we can use as the third letter, and so on. This means we have

$8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 8!$

ways of choosing the remaining 8 letters.

However, some of the 8 available letters are equal: we have 2 m's, 2 t's and 2 e's. This means that we overcounted the number of words on the reasoning above. We can correct the counting by dividing three times by 2!, one time for each of the repeated letters.

The final answer is

$\dfrac{8!}{2! \times 2! \times 2!} = \dfrac{40320}{8} = 5040.$

Answer 2

Using factorials, it is found that 5040 distinct words can be formed.

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• A word has n letters.
• Consider that there are m repeating letters, each repeating $n_1,n_2,...,n_m$ times
• The number of distinct words that can be formed is given by:

$N = \frac{n!}{n_1!n_2!...n_m!}$

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• Committee has 9 letters.
• We want words starting with i, thus, we consider 8 letters, thus $n! = 8$.
• m, t and e repeat twice, and thus, $n_1 = n_2 = n_3 = 2$

The number of distinct words is given by:

$N = \frac{8!}{2!2!2!} = 5040$

5040 distinct words can be formed.

A similar problem is given at https://brainly.com/question/16088294