Question -
Answer -
Given:
The word ‘PARALLEL’
There are 8 letters inthe word ‘PARALLEL’ out of which 2 are A’s, 3 are L’s and the rest all aredistinct.
So by using theformula,
n!/ (p! × q! × r!)
total number ofarrangements = 8! / (2! 3!)
= [8×7×6×5×4×3×2×1] /(2×1×3×2×1)
= 8×7×5×4×3×1
= 3360
Now, let us considerall L’s together as one letter, so we have 6 letters out of which A repeats 2times and others are distinct.
These 6 letters can bearranged in 6! / 2! Ways.
The number of words inwhich all L’s come together = 6! / 2!
= [6×5×4×3×2×1] / (2×1)
= 6×5×4×3
= 360
So, now the number ofwords in which all L’s do not come together = total number of arrangements –The number of words in which all L’s come together
= 3360 – 360 = 3000