Find the number of permutations of letters in each of the following wo

1.	Find the number of permutations of letters in each of the following words.(i) DIVYA(ii) SHANTARAM(iii) REPRESENT(iv) COMBINE(v) BAL BHARATI
Answer» (i) There are 5 distinct letters in the word DIVYA. ∴ Number of permutations of the letters of the word DIVYA = 5! = 120 (ii) There are 9 letters in the word SHANTARAM in which ‘A’ is repeated 3 times. ∴ Number of permutations of the letters of the word SHANTARAM = \(\frac{9!}{3!}\) = 9 x 8 x 7 x 6 x 5 x 4 = 60480 (iii) There are 9 letters in the word REPRESENT in which ‘E’ is repeated 3 times and ‘R’ is repeated 2 times. ∴ Number of permutations of the letters of the word REPRESENT = \(\frac{9!}{3!2!}\) = \(\frac{9\times8\times7\times6\times5\times4}2\) = 30240 (iv) There are 7 distinct letters in the word COMBINE. ∴ Number of permutations of the letters of the word COMBINE = 7! = 5040 (v) There are 10 letters in the word BALBHARATI in which ‘B’ is repeated 2 times and ‘A’ is repeated 3 times. ∴ Number of permutations of the letters of the word BALBHARATI = \(\frac{10!}{2!3!}\) \(=\frac{10\times9\times8\times7\times6\times5\times4\times3\times2\times1}{2\times3\times2}\) = 302400

Find the number of permutations of letters in each of the following words.(i) DIVYA(ii) SHANTARAM(iii) REPRESENT(iv) COMBINE(v) BAL BHARATI

Answer»

(i) There are 5 distinct letters in the word DIVYA.

∴ Number of permutations of the letters of the word DIVYA = 5! = 120

(ii) There are 9 letters in the word SHANTARAM in which ‘A’ is repeated 3 times.

∴ Number of permutations of the letters of the word SHANTARAM = \(\frac{9!}{3!}\)

= 9 x 8 x 7 x 6 x 5 x 4

= 60480

(iii) There are 9 letters in the word REPRESENT in which ‘E’ is repeated 3 times and ‘R’ is repeated 2 times.

∴ Number of permutations of the letters of the word REPRESENT = \(\frac{9!}{3!2!}\)

= \(\frac{9\times8\times7\times6\times5\times4}2\) = 30240

(iv) There are 7 distinct letters in the word COMBINE.

∴ Number of permutations of the letters of the word COMBINE = 7! = 5040

(v) There are 10 letters in the word BALBHARATI in which ‘B’ is repeated 2 times and ‘A’ is repeated 3 times.

∴ Number of permutations of the letters of the word BALBHARATI = \(\frac{10!}{2!3!}\)

\(=\frac{10\times9\times8\times7\times6\times5\times4\times3\times2\times1}{2\times3\times2}\)

= 302400