(C) The word $BHARAT$ contains $6$ letters,where $A$ repeats $2$ times.Total number of arrangements $ = \frac{6!}{2!} = \frac{720}{2} = 360$.Now,consi

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(C) The word $BHARAT$ contains $6$ letters,where $A$ repeats $2$ times.Total number of arrangements $ = \frac{6!}{2!} = \frac{720}{2} = 360$.Now,consider the case where $B$ and $H$ always come together. Treat $(BH)$ as a single unit. The letters are $(BH), A, R, A, T$. There are $5$ units,where $A$ repeats $2$ times.The number of arrangements where $B$ and $H$ are together $ = \frac{5!}{2!} \times 2! = 120 \times 1 = 120$.Therefore,the number of words where $B$ and $H$ never come together $ = 360 - 120 = 240$.

Class 11 Mathematics Permutation and Combination Definition of permutation, Number of permutations with or without repetition, Conditional permutations

Medium

How many words can be formed using the letters of the word $BHARAT$ such that $B$ and $H$ never come together?

IIT 1977 All IIT

A
$360$
B
$300$
C
$240$
D
$120$

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Permutation and Combination

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Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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How many words can be formed using the letters of the word $BHARAT$ such that $B$ and $H$ never come together?

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