Combinatorics

STEVEN SPIELBERG wrote:Q: How many words can be formed by using 4 letters at a time out of all the letters of the word MATHEMATICS?

(A)2445

(B)2454

(C)1243

(D)1454

(E)4542

OA B

Letter options:
AA, C, E, H, I, MM, TT, S

Case 1: 4 different letters
Number of options for the 1st letter = 8. (Any of the 8 letters listed above.)
Number of options for the 2nd letter = 7. (Any of the 7 remaining letters.)
Number of options for the 3rd letter = 6. (Any of the 6 remaining letters.)
Number of options for the 4th letter = 5. (Any of the 5 remaining letters.)
To combine these options, we multiply:
8*7*6*5 = 1680.

Case 2: 2 letters the same, the other 2 letters different
Number of options for the 2 letters that are the same = 3. (AA, MM, or TT.)
These two letters must occupy a COMBINATION OF 2 POSITIONS in the 4-letter word.
Number of combinations of 2 that can be formed from the 4 positions in the word = 4C2 = (4*3)/(2*1) = 6.
Number of options for the next letter selected = 7. (Any of the 7 remaining letters.)
Number of options for the next letter selected = 6. (Any of the 6 remaining letters.)
To combined these options, we multiply:
3*6*7*6 = 756.

Case 3: 2 letters the same, the other 2 letters also the same
Options: AA, MM and TT.
The easiest approach might be to make a list of viable arrangments:
AAMM
AMAM
AMMA
MAAM
MAMA
MMAM
Since there are 6 distinct arrangements for AAMM, there will also be 6 distinct arrangements for AATT and for MMTT.
Total distinct arrangments = 6+6+6 = 18.

Thus:
Total options = 1680 + 756 + 18 = 2454.

The correct answer is B.

Note: This problem is good for practice but far too complex for the GMAT.

The word "mathematics" has 11 letters, however only 8 letters and distinct (AMT) are repeated.

I broke this in stages so it will be easier to understand

1) There are 8 letters to select from 8x7x6x5 = 1680 (This part is fairly simple)

2) Now since there are 3 letters that are repeated, we will have duplications when we select A(i)A(ii)M(i)M(ii) - It is because A(i)is not different from A(ii). It does not matter which A we select first

Hence we need to work out how many 4 letter combinations are there with double letters repeated - AAMM.

There are 3 letters (ATM) that are repeated and therefore 3C2 = 3 ways (Out of 3 you are selecting 2) to get double letters such as AATT, AAMM, TTMM.
Now we need to work out how many ways those repeated letters can be arranged in the form of AAMM, AMAM, AMMA, MMAA, MAMA and MAAM
That is 4 letters to be arranged using 2 repeated letters from (ATM) = 4!/2!2! = 6 arrangements

Now there are 3 ways and each has 6 arrangements, which works out to be 3 x 6 = 18

3) With the similar logic as discussed in point 2 we need to work out 4 letter words have the format thats have just one repeated value (AAME)
There are 3C1 ways you will land up with 3!/2! = 3 (Not its same as point 2 but, its just a coincidence The formulae is 3C1 )


Now we need to work out how many ways with one repeated letter can be arranged in the format as there are 8 letters and since we have already used one of them we are left with 7 letters.
7C2 = 7!/5!2! = 21 arrangements using 2 distinct letters.

Now we have 3 ways and 21 arrangements 3 x 21 = 63 formats.

This one is a little tricky now: Arrangements using 2 distinct letters CAN BE ARRANGED IN
We now have to work out the number of arrangements FOR 4 letters with 2 distinct letters = 4!(Four Letters)/2!(Repeated Letters) = 12
Assuming the set of letters was TT ES , then the 12 arrangements are TTES TTSE, TETS, TSTE, TEST, TSET, ESTT, SATT, ETST, STET, ETTS, STTE,

We get 63 x 12 = 756 combinations in total for 2 repeated letters and 2 distinct letters

The ans = 1680+18+756 = 2454

I hope it helps

1st letter: 8 different choices
2nd letter: 7 or 8 choices -> maximum = 8
3rd letter: 7 choices
4th letter: 6 or 7 choices -> maximum = 7

Maximum total = 8 * 8 * 7 * 7 = 3136

Who can see why this is wrong?