Consider all the permutations of the word "ENDEANOEL" Consider all the permutations of the word "ENDEANOEL" :
1)What is the number of permutations containing the word "ENDEA" ? 
I can't understand how to approach this problem!!
2)Number of permutations in which letter E occurs in the first and last position ?
 Ok I tried this: Out of nine letters i placed two E's at the edges.Out of the remaining two similar N's and 5 different letters present.So probably it is 7!/2!
3)The number of permutations in which none of the letters D,L,N  do not occur in the last 5 positions. This I got as 4!/2!*5!/3!
4)The number of permutations in which letters A,E,O occurs only in odd positions?
This I got as $4!/2! * 5!/3!$
So can someone please help me with the first question? Hints will suffice!And are the other results correct?
 A: The answers and their explanations are as follows:
1) E,N,D,E,A have been used so the alphabets left are N,O,E,L now they can be placed before or after the instance of ENDEA,
So the cases are ____ENDEA, ENDEA, __ENDEA,_ENDEA___,ENDEA___ each case has 4! permutations so the answer is (after taking ENDEA as one alphabet X) WE need arrangements of X,N,O,E,L so its $$5!$$
2) You are correct on this the answer is $$\frac{7!}{2!}$$ 
3) You are correct on this too since there are 9 alphabets and 4 of them can't be used for the final 5 places it can be said that $$\frac{4!}{2!} and \frac{5!}{3!}=\frac{4!*5!}{2!*3!} $$
4) You are correct on this too since there are only 5 odd places and 5 alphabets to place in the and the even places needs to be filled with the rest 4 it can be said that $$\frac{4!}{2!} and \frac{5!}{3!}=\frac{4!*5!}{2!*3!} $$
So you were right on all those you answered. Remember for future that when something needs to be a subset(a part of) something place things before and after it by making cases.
