
The numbers of words from the letters of the word ‘BHARAT’ in which B and H will never come together, is
A) \[360\]
B) \[240\]
C) \[120\]
D) none of these
Answer
528.5k+ views
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 \[n!.\]
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 \[n!.\]
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