    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  Suggest Corrections  1      Explore more