Find the numbers of words with or without meaning which can be made using all the letters of the word AGAIN. I

Exercise 6.3 · Q13

Q. Find the numbers of words with or without meaning which can be made using all the letters of the word AGAIN. If all these words are arranged as in dictionary what will be the 49th word, 50th word?

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Count distinct permutations of the multiset A,A,G,I,N, then locate the 49th/50th words by peeling off letters in dictionary order.

n

Total distinct words $=\dfrac{5!}{2!}=\dfrac{120}{2}=60$ .
Words starting with A: fix one A first; remaining letters are A, G, I, N — now all distinct (only 1 A left) — arranged in $4!=24$ ways. These occupy ranks $1$ – $24$ .
Words starting with G: fix G first; remaining letters A, A, I, N (A repeated twice) arranged in $\dfrac{4!}{2!}=12$ ways. Ranks $25$ – $36$ .
Words starting with I: fix I first; remaining A, A, G, N: $\dfrac{4!}{2!}=12$ ways. Ranks $37$ – $48$ .
Words starting with N: fix N first; remaining A, A, G, I: $\dfrac{4!}{2!}=12$ ways. Ranks $49$ – $60$ .
So the 49th and 50th words are the 1st and 2nd words within the N-block. …

Q.Find the numbers of words with or without meaning which can be made using all the letters of the word AGAIN. If all these words are arranged as in dictionary what will be the 49th word, 50th word?