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With eleven distinct consonants and five different vowels, how many distinct six letter words can be formed if middle two letters are occupied by vowels (may be repeated) and first two and last two positions are occupied by consonants (all distinct) ?

I tried Solving it:

$$11 \times 10 \times 5 \times 5 \times 9 \times 8 = 198,000$$

Is this the right approach? I am confused because question says vowels may be repeated.. so do we have to consider two different cases 1) with repetition and 2) without repetition?

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  • $\begingroup$ Did you mean six letter words? $198,000$? $\endgroup$ – N. F. Taussig Oct 9 '16 at 12:18
  • $\begingroup$ yes, distinct six letter word! $\endgroup$ – Rahul Singh Oct 9 '16 at 12:19
  • $\begingroup$ Your solution is correct. $\endgroup$ – N. F. Taussig Oct 9 '16 at 12:25
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Yes, this is correct.

Let's look at the number of ways one can do this:

Position 1: $11$ options

Position 2: $10$ options (one has been used up already)

Position 3: $5$ options

Position 4: $5$ options (one hasn't been "used up" this time)

Position 5: $9$ options (continuing from position 2)

Position 6: $8$ options

And thus we have

$$11\cdot10\cdot5\cdot5\cdot9\cdot8=198000$$

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