How many $3$-character strings can be formed using the letters of word MISSISSIPPI? Only the letters that repeat in the original word can be repeated. I've tried $\frac{11!}{4!4!2!}$ and the for n = then result of the previous permutation I did $\binom{n}{3}$. Just feels like there are way more values than necessary.
 A: Hint
Since you don't know exponential g.f's, an easy  mechanical alternative is outlined below:
There are $4 S's, 4 I's, 2 P's \;and\; 1 M$
Form $4$ digit numbers from largest to smallest,  (from the characters available),  each row summing to $3$, eg
$3000\;2100\;2010\;2001\;1200\;1110 ...\;$ and add their permutations, viz
$3!(\frac 1{3!} +\frac1{2!1!} +\frac 1{2!1!} + \frac1{2!1!}+\frac1{1!2!}+\frac1{1!1!1!}...)$
A: Answer:
We see that there is 1 M, 4 I's, 4 S's, and 2P's in MISSISSIPPI.
As we want 3 letters, we, technically, have 1 M, 3 I's, 3 S's and 2P's now.
We have MIS, MIP, MSP, MSI, MPI, MPS, MII, MSS, MPP - 9 for M

 IMS,ISM,IMP,IPM,ISP,IPS,ISS,IPP,IIM,IIS,IIP,III,IMI,IPI,ISI - 15 for I, and 15 again for S.


 PMI,PIM,PMS,PMP,PIP,PSP,PSM,PIS,PSI,PII,PSS,PPM,PPI,PPS - 14.


 Therefore, our answer will be $9+15+15+14=53$.

A: Rather than coming up with a single expression, it is easier to enumerate different possibilities. We can start with M: either there is one, or there isn't.

 If there is no M, the three characters can be any combination of I, S and P, with just one exception: it can't be PPP. So there are $3^3 - 1 = 26$ possibilities.


 If there is a M, then there are three possibilities for its position. The other two characters can be any combination of I, S and P. So there are $3 \times 3^2 = 27$ possibilities.


 Conclusion: in total there are $26 + 27 = 53$ possibilities.

A: The first letter can be any of 4 possibilities (I, M, P, or S).
The second letter can also be any of those four except that MM is not a valid combination (as MISSISSIPPI has only one M).  So there are $4 \times 4-1=15$ valid permutations for the first two letters.
Having the same 4 options for the third letter brings us up to $15 \times 4 = 60$ permutations, but we must exclude PPP (since we're limited to two P's), and six permutations that already have an M in the first two letters (IMM, MIM, MPM, MSM, PMM, and SMM).
This leaves us with 53 valid permutations: III, IIM, IIP, IIS, IMI, IMP, IMS, IPI, IPM, IPP, IPS, ISI, ISM, ISP, ISS, MII, MIP, MIS, MPI, MPP, MPS, MSI, MSP, MSS, PII, PIM, PIP, PIS, PMI, PMP, PMS, PPI, PPM, PPS, PSI, PSM, PSP, PSS, SII, SIM, SIP, SIS, SMI, SMP, SMS, SPI, SPM, SPP, SPS, SSI, SSM, SSP, and SSS.
