Any answer for this guys....?

Hi Sudhir,

Where did you get this problem? I don't think selections is an official way to say combinations. It is somewhat misleading. Also, there are only 4 answer choices to the first problem, and no answer choices to the second, which makes me wonder if this problem came from somewhere besides a reputable GMAT source.... I worked out your problem but did not come up with any of the numbers given by your answer choices. Are you positive about these answers?

Tatiana

Any one for thi question .. its from a genuine source...

Not sure if I have understood the question completely. As per my understanding of the question "we are looking to see the possible combinations of 4 letters from the word"ENTRANCE".

So ENTRANCE has 6 unique letters.

So the ANS should be 6C4=6C2=15

But I don't see this in answer choices.
Any comments.

These questions are really beautiful. It would be nice if you could refer the name of the book or source. I definitely dont doubt the credibility of the question, these question can appear on the actual GMAT. These questions are GOLDMINE!!!!

ENTRANCE

1) The first problem is of combination.

4 letters are to be chosen out of 8 letters but 2 letters are in repetition.

There are 6 distinct letters. Lets take this case by case.

I......... 4 letters are chosen from 6 distinct letters.

6C4 = 15 ways

II........E & N are in repetition therefore only way we can choose both of them together i.e. NN EE is 1 way

III......No. of ways 2 EE's are chosen in 4 letter is 1*5C2 = 10 ways
(1*5C2 means we can put 2 E in either of the 2 letters of 4 letters and remaining we are left with 2 more letters to be chosen out of 5 distinct letters, therefore 5C2)

IV......No. of ways 2 NN's are chosen in 4 letter is 1*5C2 = 10 ways


Therefore total number of selections is 15+1+10+10= 36


2) The second question is bit tricky. If you understand the anagram form this can be a bit easier to understand.

I.....lets take only distinct letters i.e. ENTRAC (6 letters)

These can be arranged in 6*5*4*3 or 6P4 = 360 ways.

II....lets take the repetitions EENN
These can be arranged in 4!/(2!2!) = 6 ways

III...lets take condition only EE and TRANC (1 repetition + 5 distinct letters)
12/2! * 5*4 = 120 ways or 12!/2*5P2

12 because in first two places we can arrange one E in 4 ways and the other E in 3 ways but since we cannot differentiate between the two EE we have to divide by 2. IF WE DONT DIVIDE BY 2 THE METHOD WILL DIFFERENTIATE E1 E2 T R & E2 E1 T R which we dont want.


IV....Condition NN and TRACE (Reasoning same as above)
12!/2 * 5*4 = 120 ways or 12!/2*5P2 = 120 ways


Add all of them 360+6+120+120 = 606 arrangements.

Another way of looking at the same problem:

How many 4 digit numbers can formed using letters 1,2,2,3,3,4,5,6.

It took me more than half hour to do these, but I still believe that these are one of the best questions I have ever solved for permutation & combination.

Thanks parallel chase...the questions are from one of the Entrance tests for Business Schools in India ( IIM ) CAT. considered to be one the toughest entrances to crack...

thanks for ur answers.....

sudhir3127 wrote:Thanks parallel chase...the questions are from one of the Entrance tests for Business Schools in India ( IIM ) CAT. considered to be one the toughest entrances to crack...

thanks for ur answers.....

No wonder these questions are from IIMs entrance, my sister sat for the exam about 7 years ago scored 99 percentile but still couldnt get through.

Anyways it'd be nice if you could post some more questions like these.

Thanks.