How many four letter words can be formed using the letterss of the word 'INEFFECTIVE'? 
How many four letter words can be formed using the letterss of the word 'INEFFECTIVE'?

Please explain with details. Also, how can we clearly differentiate between permutations and combinations?
 A: Letters:-

C-1
E-3
F-2
I-2
N-1
T-1
V-1
ABCD type words : $^7C_4\times4!$
ABCC type words : $^3C_1\times^6C_2\times\frac{4!}{2!}$
ABBB type words : $1\times^6C_1\times\frac{4!}{3!}$
AABB type words : $^3C_2\times\frac{4!}{2!\times2!}$
Sum them over to get the total!
NOTE: AABB (for example) means all words (including permutations) with 2 letters of one kind and two letters of another type.
A: There are $11$ letters in 'INEFFECTIVE'.  

CASE $(1)$: There is only set of three same letters of the letter $E$. So $3$ same letters can be selected in one way. Out of the six remaining distinct letters we can select one in $\binom{6}{1}=6$ ways. Hence there are $6$ groups each of which contains the three same letters. These four letters can be arranged in $\binom{4}{3}=4$ ways. Hence the total number of words with $3$ same and one different letters is $6\times 4=24$ ways.

CASE $(2)$: There are $3$ sets of two same letters $E,F,I$. Out of these $3$ sets, two can be selected in $\binom{3}{2}=3$ ways. So there are $3$ groups each of which contains $4$ letters two of which are same of one type and the other two of a different same type. Now $4$ letters in each group can be arranged in $\binom{4}{2}=6$ ways. Hence the total number of words with two letters of one kind and the other two of the different kind is $3\times 6=18$ times.  

CASE $(3)$: Out of $3$ sets of $2$ same letters one set can be chosen in $\binom{3}{1}=3$ ways. Now from the remaining $6$ letters, $2$ distinct letters can be chosen in $\binom{6}{2}=15$ ways. So there are $3\times 15=45$ groups of $4$ letters each. Now letters of each group can be arranged in $\frac{4!}{2!}=12$ ways. Hence, the total number of ways consisting of two same letters and $2$ different are $45\times 12=540$ ways.  

CASE $(4)$: There are $\binom{7}{4}$ groups of words with $4$ different letters each. In each group each letter can be arranged in $4!=24$ ways. So the total number of $4$ letter words in which all letters are different is $\binom{7}{4}\times 24=840$ ways.  

There are thus a total of $24+18+540+840=1422$ ways.

