  Question

Consider all possible permutations of the letters of the word ENDEANOEL.

Solution

Nine letters : $$3E^{s}, 2N^{s}, 4$$ different $$D, A, O, L$$.$$ENDEA, N,O, E, L$$ Five different$$(a)\rightarrow (p)$$ String Method, We have five different units whose permutation is $$5!$$.$$(b)\rightarrow (s) E$$ ENNDAOL $$E$$$$1st$$ and last places are filled with $$E^{S}$$. We have to arrange remaining $$7$$ letters out of which $$2N^{S}$$ are alike.$$\therefore \dfrac {7!}{2!} = \dfrac {7 . 6. 5!}{2} = 21(5!)$$$$(c)\rightarrow (q) D, L, N, N$$ will not occur in last five and hence they have to be arranged in first $$4$$ position. $$\therefore \dfrac {4!}{2!}$$For the last five letters out of which $$3E^{S}$$ are alike $$\dfrac {5!}{3!}$$.Hence total $$= \dfrac {4!}{2!}\cdot \dfrac {5!}{3!} = \dfrac {4.(3!)}{2}\cdot \dfrac {5!}{3!} = 2(5!)$$$$(d)\rightarrow (q) AEO$$ to occur in odd positions. i.e., $$1st, 3rd, 5th, 7th$$ and $$9th$$.$$\dfrac {5!}{3!(alike E)} \cdot \dfrac {4!}{2!(alike N)} = 5! \dfrac {4}{2} = 2(5!)$$Maths

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