Question -
Answer -
(i) INDEPENDENCE
There are 12 lettersin the word ‘INDEPENDENCE’ out of which 2 are D’s, 3 are N’s, 4 are E’s and therest all are distinct.
So by using theformula,
n!/ (p! × q! × r!)
total number ofarrangements = 12! / (2! 3! 4!)
=[12×11×10×9×8×7×6×5×4×3×2×1] / (2! 3! 4!)
= [12×11×10×9×8×7×6×5]/ (2×1×3×2×1)
= 11×10×9×8×7×6×5
= 1663200
(ii) INTERMEDIATE
There are 12 lettersin the word ‘INTERMEDIATE’ out of which 2 are I’s, 2 are T’s, 3 are E’s and therest all are distinct.
So by using theformula,
n!/ (p! × q! × r!)
total number ofarrangements = 12! / (2! 2! 3!)
= [12×11×10×9×8×7×6×5×4×3×2×1]/ (2! 2! 3!)
=[12×11×10×9×8×7×6×5×3×2×1] / (3!)
= 12×11×10×9×8×7×6×5
= 19958400
(iii) ARRANGE
There are 7 letters inthe word ‘ARRANGE’ out of which 2 are A’s, 2 are R’s and the rest all aredistinct.
So by using theformula,
n!/ (p! × q! × r!)
total number ofarrangements = 7! / (2! 2!)
= [7×6×5×4×3×2×1] /(2! 2!)
= 7×6×5×3×2×1
= 1260
(iv) INDIA
There are 5 letters inthe word ‘INDIA’ out of which 2 are I’s and the rest all are distinct.
So by using theformula,
n!/ (p! × q! × r!)
total number ofarrangements = 5! / (2!)
= [5×4×3×2×1] / 2!
= 5×4×3
= 60
(v) PAKISTAN
There are 8 letters inthe word ‘PAKISTAN’ out of which 2 are A’s and the rest all are distinct.
So by using theformula,
n!/ (p! × q! × r!)
total number ofarrangements = 8! / (2!)
= [8×7×6×5×4×3×2×1] /2!
= 8×7×6×5×4×3
= 20160
(vi) RUSSIA
There are 6 letters inthe word ‘RUSSIA’ out of which 2 are S’s and the rest all are distinct.
So by using theformula,
n!/ (p! × q! × r!)
total number ofarrangements = 6! / (2!)
= [6×5×4×3×2×1] / 2!
= 6×5×4×3
= 360
(vii) SERIES
There are 6 letters inthe word ‘SERIES’ out of which 2 are S’s, 2 are E’s and the rest all aredistinct.
So by using theformula,
n!/ (p! × q! × r!)
total number ofarrangements = 6! / (2! 2!)
= [6×5×4×3×2×1] / (2!2!)
= 6×5×3×2×1
= 180
(viii) EXERCISES
There are 9 letters inthe word ‘EXERCISES’ out of which 3 are E’s, 2 are S’s and the rest all aredistinct.
So by using theformula,
n!/ (p! × q! × r!)
total number ofarrangements = 9! / (3! 2!)
= [9×8×7×6×5×4×3×2×1]/ (3! 2!)
= [9×8×7×6×5×4×3×2×1]/ (3×2×1×2×1)
= 9×8×7×5×4×3×1
= 30240
(ix) CONSTANTINOPLE
There are 14 lettersin the word ‘CONSTANTINOPLE’ out of which 2 are O’s, 3 are N’s, 2 are T’s andthe rest all are distinct.
So by using theformula,
n!/ (p! × q! × r!)
total number ofarrangements = 14! / (2! 3! 2!)
= 14!/ (2×1×3×2×1×2×1)
= 14! / 24