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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
Verified
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
Is there an error in this question or solution?
APPEARS IN
How many different words can be formed with the word Bharat?
How many words with or without meaning can be formed from the letter Bharat?
How many different words can be formed using the letters of the word Bharat II how many words begin with B and end with T?
How many permutations of the letters of the word INDIA are there?
Number of permutations of the letters of the given word `=[5!]/[2!]= 60.