Question

Find the number of words formed by permuting all the letters of the following words:

(i) INDEPENDENCE

(ii) INTERMEDIATE

(iii) ARRANGE

(iv) INDIA

(v) PAKISTAN

(vi) RUSSIA

(vii) SERIES

(viii) EXERCISES

(ix) CONSTANTINOPLE

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Solution

There are 12 letters in the word `INDEPENDENCE' out of which 2 are D'S, 3 are N'S, 4 are E'S and the rest are all distinct.

So, the total number of words = 12!2!3!4!

= 12×11×10×9×8×7×6×5×4!2!3!4!

= 12×11×10×9×8×8×7×6×52×3×2

= 11×10×9×8×7×6×5

= 1663200

(ii) There are 12 letters in the word 'INTERMEDIATE' out of which 2 are I'S, 2 are T'S, 3 are E'S and the rest are all distinct.

So, the total number of words

= 12!2!2!3!

=12×11×10×9×8×7×6×5×4×3×2!2!2!3!

= 12×11×10×9×8×7×6×5×4×32×3×2

= 11×10×9×8×7×6×5×4×3

= 19958400

(iii) There are 7 letters in the word 'ARRANGE'out of which 2 are A'S, 2 are R'S, and the rest are all distinct.

So, the total number of words = 7!2!2!

= 7×6×5×4×3×2!2!2!

= 7×6×5times4×32×1

=1260

(iv) There are 5 letters in the word 'INDIA' out of which 2 are I'S, and the rest are all distinct.

So, the total number of words = 5!2!

= 5×4×3×22!

= 60

(v) There are 8 letters in the word 'PAKISTAN' out of which 2 are A'S, and the rest are all distinct.

So, the total number of words

=8!2!

= 8×7×6×5×4×3×2!2!

= 8×7×6×5×4×3 = 20160

(vi) There are 6 letters in the rest are all distinct.

So, the total number of words

= 6!2!

= 6×5×4×3×2!2!

= 6×5×4×3=360

(vii) There are 6 letters in the word 'SERIES' out of which 2 are S's, 2 are E's and the rest are all distinct.

So, the total number of words

= 6!2!2!

= 6×5×4×3×2!2!2!

= 6×5×4×32×1

= 6×5×2×3=180

(viii) There are 9 letters in the word 'EXERCISES' out of which 3 are E's, 2 are S's and the rest are all distinct.

So, the total number of words

words = 14!2!3!2!

= 14!2×3×2×2

= 14!2×3×2×2×2

= 14!24

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