How many words can be formed by rearranging the alphabets of word 'REH

1.	How many words can be formed by rearranging the alphabets of word 'REHABILITATION'? (must not necessarily have meaning)1. \(\rm 14!\over2!\times2!\)2. \(\rm 14!\over2!\times3!\)3. \(\rm 14!\over2!\times2!\times3!\)4. \(\rm 14!\over4!\times3!\)
Answer» Correct Answer - Option 3 : \(\rm 14!\over2!\times2!\times3!\) Concept: The ways of arranging n different things = n! The ways of arranging n things, having r same things and rest all are different = \(\rm n!\over r!\) The no. of ways of arranging the n arranged thing and m arranged things together = n! × m! The number of ways for selecting r from a group of n (n > r) = nCr Calculation: R E H A₁ B I₁ L I₂ T₁ A₂ T₂ I₃ O N Total of letters = 14 There are 2 A's, 3 I's, 2 T's, ∴ The number of words formed = \(\rm 14!\over2!\times2!\times3!\)

How many words can be formed by rearranging the alphabets of word 'REHABILITATION'? (must not necessarily have meaning)1. \(\rm 14!\over2!\times2!\)2. \(\rm 14!\over2!\times3!\)3. \(\rm 14!\over2!\times2!\times3!\)4. \(\rm 14!\over4!\times3!\)

Answer» Correct Answer - Option 3 : \(\rm 14!\over2!\times2!\times3!\)

Concept:

The ways of arranging n different things = n!
The ways of arranging n things, having r same things and rest all are different = \(\rm n!\over r!\)
The no. of ways of arranging the n arranged thing and m arranged things together = n! × m!
The number of ways for selecting r from a group of n (n > r) = nCr

Calculation:

R E H A₁ B I₁ L I₂ T₁ A₂ T₂ I₃ O N

Total of letters = 14

There are 2 A's, 3 I's, 2 T's,

∴ The number of words formed = \(\rm 14!\over2!\times2!\times3!\)

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How many words can be formed by rearranging the alphabets of word 'REHABILITATION'? (must not necessarily have meaning)1. \(\rm 14!\over2!\times2!\)2. \(\rm 14!\over2!\times3!\)3. \(\rm 14!\over2!\times2!\times3!\)4. \(\rm 14!\over4!\times3!\)

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