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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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