In how many ways can we arrange 4 letters of the word "ENGINE"? I need to know a combinatoric solution to this problem, with Generating functions in book, gives us 102. 
It might be a very simple problem, but Im very confused with this.
 Would be very nice of you to give me some help. 
Im really grateful.
 A: We can view the letters of the word ENGINE as a multiset with two E's, two N's, one G, and one I, that is $\{2 \cdot E, 2 \cdot I, 1 \cdot G, 1 \cdot I\}$.  When we select four of these letters, we will either have four different letters, three different letters with one repeat, or two different letters with each repeated.  
Case 1:  Four different letters. 
This is a permutation of the letters E, N, G, I.  They can be permuted in $4!$ ways.
Case 2:  Three different letters with one repeated.
There are two ways to select the repeated letter from $\{E, N\}$.  We can choose the locations of the repeated letter in $\binom{4}{2}$ ways.  There are three ways to fill the leftmost open spot with one of the other three letters and two ways to fill the last spot with one of the two letters that has not yet been selected.  Hence, there are 
$$\binom{2}{1}\binom{4}{2}\binom{3}{1}\binom{2}{1}$$
arrangements with three different letters in which one letter is repeated.
Case 3:  Two different letters with both repeated.
We permute two E's and two N's.  There are $\binom{4}{2}$ ways to select the locations of the two E's.  There is one way to fill the remaining spots with the two N's.  Hence, the number of ways to arrange two different letters with both repeated is
$$\binom{4}{2}$$
Thus, the total number of ways of arranging four letters of the word ENGINE is 
$$4! + \binom{2}{1}\binom{4}{2}\binom{3}{1}\binom{2}{1} + \binom{4}{2}$$
A: Add up the following:


*

*The number of ways to arrange EEGI: $\frac{4!}{2!\cdot1!\cdot1!       }=12$

*The number of ways to arrange EEGN: $\frac{4!}{2!\cdot1!\cdot1!       }=12$

*The number of ways to arrange EEIN: $\frac{4!}{2!\cdot1!\cdot1!       }=12$

*The number of ways to arrange EENN: $\frac{4!}{2!\cdot2!              }= 6$

*The number of ways to arrange EGIN: $\frac{4!}{1!\cdot1!\cdot1!\cdot1!}=24$

*The number of ways to arrange EGNN: $\frac{4!}{1!\cdot1!\cdot2!       }=12$

*The number of ways to arrange EINN: $\frac{4!}{1!\cdot1!\cdot2!       }=12$

*The number of ways to arrange GINN: $\frac{4!}{1!\cdot1!\cdot2!       }=12$

A: There are 3 cases in for this 
Case1: When all the 4 letters differ 
Solution:
Since we have only 4 different letters in the word ENGINE
 = 4! = 24
       =24

Case2: When a single letter is repeated twice
Solution:
Since there are two letter that can be selected twice (i.e EE & NN)
We can select a set in 2C1 = 2 (we want a set out of the 2 sets)
And the remaining 2 letters can be selected from a total of  3 different letters
Therefore it's 3C2 = 3
But in how many ways can we arrange all this? 
I.e 4!/2! = 12 Since there are two repeated numbers 
      2 × 3 × 12 = 72

Case 3: When two distinct letters are repeated twice each
Since there are only 2Ns and 2Es 
This can be arranged in 4!/2!×2! = 6 Since two letters are repeated 
        = 6

Overall 
       24 + 72 + 6 = 21

A: By grouping the similar things we got,
G, I, (E, E), (N, N)

*

*4 letters are different, then we can select =
4C4, arrange = 4C4 × 4!


*

*

*2 E and 2 letters are different, then we can select =    3C2, arrange = 3C2 × $\frac{4!}{2!       }$


*2 N and 2 letters are different, then we can select =    3C2, arrange = 3C2 × $\frac{4!}{2!}$




*Taking 2 from same group and 2 from other same group, this can be
selected in only 1 way - (for example, 2 E, 2 N), we can select = 1,
arrange = 1 × $\frac{4!}{2! 2!      }$
By adding,
Total number of selection = 8 
Total number of arrangement = 102
