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Find the number of permutations of the letters of the “COMMITTEE”. (a) How many of them begin with T and end with T.(b) In how many all the vowels are together. (c) In how many no two vowels are together. (d) In how many of them end with MITE. |
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Answer» Total letters = 9, M = 2, T = 2, E = 2 Total number of permutations \(\frac{9!}{(2!)^3}\) (a) Number of permutations which being in with T and end with T is given by \(\frac{7!}{(2!)^2}\) (b) Vowels are together can be taken as 1 unit. 4 vowels = 1 and remaining 5, totally 6 can be permuted (I 0 E)(112) in \(\frac{6!}{(2!)^2}\) and the vowels they themselves can be done in \(\frac{4!}{(2!)}\) ways ∴ Total number of ways \(\frac{6!}{(2!)^2}\) . \(\frac{4!}{(2!)}\) (c) No two vowels are together: Should be in between the consonants, i.e. -C – M – M – T – T. – There are 6 vacant places in between consonants to arrange 4 vowels. This can be arranged in \(\frac{^6P_4}{2!}\) and 5 consonants can be arranged in \(\frac{5!}{(2!)^2}\) ∴ Total number of ways = \(\frac{^6P_4}{2!}\) x \(\frac{5!}{(2!)^2}\) (d) End with MITE: The remaining 5 letters can be arranged in 5! ways. |
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