How many different words can be formed taking 4 alphabets at

Question

How many different words can be formed taking 4 alphabets at a time from the word "INTEMPERATE"?

a) 1680
b) 1968
c) 1766
d) 2218

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BrentGMATPrepNow · Answer

This question requires us to consider wayyyyyyyy too many different cases to be an actual GMAT question.

Mugdho · Answer

BrentGMATPrepNow 
yes...but this can be good exercise...

IanStewart · Answer

I agree with Brent -- it's not a realistic problem, but we can actually get away with considering just two cases. We have 8 different letters. So if we use each letter once, we'll have 8 choices for the first letter, 7 for the second, and so on, for 8*7*6*5 = 1680 words we can make using four different letters. Most of the remaining words will look like EIEM, where we use exactly one letter twice. We have 2 choices for that letter (only T and E are repeated), and then we have 4C2 = 6 choices for which two positions those letters will occupy in the word (choosing from the 1st, 2nd, 3rd and 4th letters in the word). So we have 2*6 = 12 ways to position two repeated letters in a word, and then we'll have 7 choices for the first empty slot in the word, and 6 for the second, for 12*7*6 = 504 possible words that use one letter twice and two letters once.

We're already well past 2000 possible words, leaving D as the only plausible answer, and we don't need to count the two remaining cases (where we use both the T and E twice, which gives only 6 possibilities, and where we use the E all three times, and some other letter once, which turns out to give only 28 possibilities).

But I think it's better to practice questions more similar to what will show up on the actual test.

We're already well past 2000 possible words, leaving D as the only plausible answer, and we don't need to count the two remaining cases (where we use both the T and E twice, which gives only 6 possibilities, and where we use the E all three times, and some other letter once, which turns out to give only 28 possibilities).

But I think it's better to practice questions more similar to what will show up on the actual test.

PyjamaScientist · Answer

Totally agree with  BrentGMATPrepNow  and  IanStewart  above.
 Mugdho  even if you posted this for practice purposes, you still got the OA wrong. C in no way can be the answer. And through my "practice" after considering all the possible scenarios, I got 2218! That is option D. So, I highly recommend you to at least change the OA even if it's just for practice. Because, to see that red light pop up once you click on an answer choice is a bummer!

lordwells · Answer

Okay, here's my solution
Unfortunately I already know it is wrong since it doesn't fit the OA, but I would like to know why it is the case:
So we are essentially given
11 letters
8 of which are unique and 2 are repeated "E" and 1 repeated "T"
Different possible scenarios are:
1) composing a 4-letter word using only non-repeated letters
8C4 * 4! = 8*7*6*5 = 1.680
2) composing a 4-letter word using a couple of repeated letter (either 2 "E" OR 2 "T")
7C2 * 4!/2! * 2 = 7*6*12 = 504
3) composing a 4-letter word using 3 "E"
7C1 * 4!/3! = 28
4) composing a 4-letter word using 2 "E" AND 2 "T"
4!/2!*2! = 6
Total = 2.218
Therefore Option D)

Where did I go wrong?

IanStewart · Answer

Your solution is perfect -- the OA recorded for the question is wrong.

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