Number of ways selecting 4 letter words 
The number of ways of selecting 4 letters out of the letters MANIMAL

A. 16
  B. 17
  C. 18
  D. 19
I have made three different cases. Including 1 M, 2 M and none of the M.  
So it is 6C3/2 + 5C3 + 4C3 which doesn't meet any of the option. 
 A: Hint
$$(1+x+x^2)^2(1+x)^3=x^7+5 x^6+12 x^5+\color{red}{18} x^4+{18} x^3+12 x^2+5 x+1$$
A: Add up the following, and you get $18$:


*

*The number of combinations with $A=0$ and $M=1$ is $\binom33=1$

*The number of combinations with $A=0$ and $M=2$ is $\binom32=3$

*The number of combinations with $A=1$ and $M=0$ is $\binom33=1$

*The number of combinations with $A=1$ and $M=1$ is $\binom32=3$

*The number of combinations with $A=1$ and $M=2$ is $\binom31=3$

*The number of combinations with $A=2$ and $M=0$ is $\binom32=3$

*The number of combinations with $A=2$ and $M=1$ is $\binom31=3$

*The number of combinations with $A=2$ and $M=2$ is $\binom30=1$

A: the answer  should be for MNIMAMAL.,
$(1+x)^3(1+x+x^2)(1+x+x^2+x^3)$ $=$ $x^8+5x^7+12x^6+19x^5+22x^4+19x^3+12x^2+5x+1$
hence,the answer will be 22
A: We can consider the number of pairs of repeated letters:
1) If there are no repeated letters, we have $\dbinom{5}{4}=5$ selections.
2) If there is one pair of repeated letters, we have $\dbinom{2}{1}\dbinom{4}{2}=12$ selections.
3) If there are 2 pairs of repeated letters, we have only 1 selection.
Therefore there are 18 possible selections altogether.
