Combination + Permutation = Combutation

The letters given are M, M, A, A, T, T, H, I, C, S, E.

There are three pairs of repeated letters.

First we shall calculate number of combinations,

Combinations:

There are three possible cases:

1. No three of the four letters in the word are repeated.

2. Only one pair of letters in the word are repeated.

3. Two pairs of letters in the word are repeated.

Case 1 : We have 8 letters namely M, A, T, H, I, C, S, E.

The number of ways of selecting 4 letters from 8 letters is $\binom{8}{4} = 70$ .

Case 2: We have three of pairs of repeating letters. We have to select one from them. This is possible in $3$ ways. For suppose to make this case even more clear we say that we select the pair M, M . We still have to select two more letters. There are seven more letters (non-repeating) namely A, T, H, I, C, S, E. So obviously there are $\binom{7}{2}$ ways.

So, total number of ways in this cases are $3 \times \binom{7}{2} = 63$ .

Case 3: We have three pairs of repeating letters and we have to select two pairs from them and this can be done in $\binom{3}{2} = 3$ ways.

So, the total number of combinations $= 136$ . Therefore, $\boxed{C=136}$ .

Permutations:

There are three possible cases just as mentioned above.

Case 1: We have 8 letters namely M, A, T, H, I, C, S, E.

The number of ways of permuting 4 letters from 8 letters is $^8 P _{4} = 1680$ .

Case 2: Similar to the case (ii) in combinations. The number of ways of selecting $4$ letters in this case is $63$ . Now we have to permute these selected four letters. The number of ways of permuting $4$ letters with $2$ letters repeated are $\dfrac{4!}{2! } = 12$ . So, total number of ways in this case is equal to $63 \times 12 = 756$ .

Case 3: The number of ways of selecting two pairs from three pairs of repeating letters is equal to $3$ . And number of ways of permuting them is equal to $\dfrac{4!}{2! 2!} = 6$ . So, the total number of ways in this case is equal to $3 \times 6 = 18$ .

So, the total number of permutations $= 2454$ . Therefore, $\boxed{P = 2454}$ .

Therefore, $P+C = \boxed{2590}$ .

I split out the possibilities based on the number of repeated letters ( M, A, T ) in the combination. The number of permutations corresponding to a given combination is easy to calculate.

In the table below, MAT stands for the number of ways to choose M, A, T in the pattern, and "other" is the number of ways to pick the remaining letters. P/C is the number of permutations for this kind of combination.

 1
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 3
 4
 5
 6
 7
 8
 9
10
11
12

(MAT) MAT other C P/C  P
===== === ==== == === ===
  -    1    5   5  24 120
  1    3   10  30  24 720
 1+1   3   10  30  24 720
  2    3   10  30  12 360
1+1+1  1    5   5  24 120
 1+2   6    5  30  12 360
 2+2   3    1   3   6  18
2+1+1  3    1   3  12  36
===== === ==== -- === ---
              136    2454

So $C = 136$ , $P = 2454$ , and their total is $\boxed{2590}$ .

Example of how to understand the table:

The row "1+2" describes combinations of the form $\langle XXYz\rangle$ , where $X,Y$ are two of M, A, T, and $z$ is a letter from H, E, I, C, S. There are $2\cdot 3 = 6$ ways to choose $X$ and $Y$ , and 5 ways to choose $z$ , so that there are $6\cdot 5 = 30$ combinations of this kind.

By the way, they are

1
2
3
4
5
6

MMAH   MMAE   MMAI   MMAC   MMAS
MMTH   MMTE   MMTI   MMTC   MMTS
AAMH   AAME   AAMI   AAMC   AAMS
AATH   AATE   AATI   AATC   AATS
TTMH   TTME   TTMI   TTMC   TTMS
TTAH   TTAE   TTAI   TTAC   TTAS

For each of these combinations there are $4!/2 = 12$ permutations. For example, the combination MMAH results in the permutations

1
2

MMAH   MMHA   MAMH   MHMA   MAHM   MHAM
AMMH   HMMA   AMHM   HMAM   AHMM   HAMM

Thus for the pattern "1+2" there are 30 combinations with 12 permutations corresponding to each combination, for a total of $30\cdot 12 = 360$ permutations.

We basically have 8 different alphabets as MM, AA, TT, H, E, I, C, S

Note that selection means combination & arrangement meant permutation.

Case 1: 2 alike, 2 alike

We have 3 '2 alike' pairs namely MM, AA, TT. We will select any two in $^{3}C _{2}$ ways and then arrange in 4!/(2!2!) ways.

So Combinations=3

Permutations=3x24/4=18

Case 2: 2 alike, 2 different

We will select any one of '2 alike' pairs in $^{3}C _{1}$ ways & 2 different letters in $^{7}C_{2}$ ways and then arrange in 4!/(2!) ways.

So Combinations=3x21=63

Permutations=63x12=756

Case 3: All 4 different

We will select 4 different alphabets in $^{8}C _{4}$ ways and then arrange in 4! ways.

So Combinations=70

Permutations=70x24=1680

From above, total P+C=2590

I arranged it as M M A A T T H E I C S using Excel.

No easy seemingly. I listed out the 136 combinations and then calculated for 6, 12 or 24 which sum up to 2454 permutations. The wording of the question should make "C be the number of combinations and P be the number of permutations" of four alphabets taken from the word MATHEMATICS, despite a clarification of Combination + Permutation = Combutation, to avoid doubt of ambiguity as the word Combutation is not found in usual website. Meaning to say, the "four taken" should be back distributed to front for its meaning wanted.

1) M M A A; 6

2) M M A T; 12

3) M M A H; 12

4) M M A E; 12

5) M M A I; 12

6) M M A C; 12

7) M M A S; 12

...

131) T I C S; 24

132) H E I C; 24

133) H E I S; 24

134) H E C S; 24

135) H I C S; 24

136) E I C S; 24

My way of listing makes a very clear situation of which imagination required is less and the confidence of correctness is more direct. The weakness is more work is required comparatively.

136 + 2454 = 2590.