Using permutation principles, we calculate the total permutations of 'bharat' as 6!, then find permutations with b and h together as (5! * 2) / 2!, and subtract the latter from the former to find permutations with b and h never together. None of the provided options match the correct answer, which is 480.
To solve the number of words that can be made from the letters of the word bharat without the letters b and h coming together, we can use the concept of permutations and combinations. First, we find the total number of ways to arrange all the letters in the word, which is the factorial of the number of letters. Since 'bharat' has 6 letters, the total permutations are 6!. Then, we subtract the permutations where b and h are together.
To find permutations where b and h are together, we treat them as one unit. So, we have now 5 units – (bh), a, r, a, t. There are 5! ways to arrange these 5 units. Since 'a' occurs twice, we need to correct for these repeat permutations, dividing by 2!. Now we multiply the 5! by 2 (two arrangements for each - bh and hb). Finally, subtract these from the total permutations to find our answer:
- Total permutations = 6! = 720
- Permutations where b and h are together = (5! * 2) / 2! = 240
- Permutations where b and h are never together = 720 - 240 = 480
So, the correct answer is D. None of these, since the closest option offered was 360, which is incorrect.