How many arrangements of the letters of the word bharat will not have b and h together-

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P and C Permutations Assignment

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the problem statement number but from the letter word Bharat in DNH will never come together we have let so we have had Marathi haran6 letter word auto 6 1 twice between two never come together and never come together to this can be arranged

DM come together total total arrangement Sobi Hai 6 letter word out of 6 letter to download a minus b n HD a single entity we are now left in a are against contrast to now Bharat wycombe hp11 second third and later but out of to come to a point wise DNH can also be represented as it reaches the end and HD x

Thoothukudi Airport Road Ahmedabad - Road 12030 into 12 - 1 2013 360 - 120 so this is equal to 240

Answer

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Hint: To solve this question, we will start with finding the total number of words formed using given letters, then we will find the number of words formed in which B and H are together, then on taking difference we will get the number of words in which B and H will never come together.

Complete step-by-step answer:
We have been given a word ‘BHARAT’ and we need to find the number of words from the letters of the word in which B and H will never come together.
So, the total number of letters in the word ‘BHARAT’ \[ = {\text{ }}6\]
We can see that in the given word, the letter "A" is repeated twice.
The number of ways of arranging ‘n’ objects where p is of one type, q is of second type, r is of third type, etc. $ = \dfrac{{n!}}{{p!q!r!....}} $
\[\therefore \] Total number of different words formed \[ = \dfrac{{6!}}{{2!}} = \dfrac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} = 360\]
Now, when B and H are together, we will treat them as single letter, then we get \[5\] letters, in which ‘A’ is repeated twice.
So, number of ways of arrangement in which when B and H are together where ‘A’ is repeated twice $ = \dfrac{{5!}}{{2!}} $
But B and H can be arranged in \[2!\] ways in themselves.
\[\therefore \]Number of arrangements with B and H
together $ = \dfrac{{5!}}{{2!}} \times 2! = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} \times 2 \times 1 = 120 $
Now, the number of words in which B and H are never together \[ = \] total number of words formed \[ - \] number of words in which B and H are together.
Number of words in which B and H are never together \[ = {\text{ }}360{\text{ }} - {\text{ }}120{\text{ }} = {\text{ }}240\]
Thus, option (B), \[240\]is correct.
So, the correct answer is “Option B”.

Note: In permutation and combination, for number of ways of arranging ‘n’ unlike object we use the formula \[n!.\]

The number of words from the letters of the word 'BHARAT' in which B and H will never come together, is

Options

• 360

• 240

• 120

• none of these.

Solution

 240
Total number of words that can be formed of the letters of the word BHARAT =\[\frac{6!}{2!}\]= 360

Number of words in which the letters B and H are always together  = \[2 \times\]\[\frac{5!}{2!}\]= 120

∴ Number of words in which the letters B and H are never together = 360 - 120 = 240

Concept: Permutations

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