Permutations And Combinations Question 42

Question: A library has $ a $ copies of one book, $ b $ copies of each of two books, $ c $ copies of each of three books and single copies of $ d $ books. The total number of ways in which these books can be distributed is

Options:

A) $ \frac{(a+b+c+d)\ !}{a\ !\ b\ !\ c\ !} $

B) $ \frac{(a+2b+3c+d)\ !}{a\ !\ {{(b\ !)}^{2}}{{(c\ !)}^{3}}} $

C) $ \frac{(a+2b+3c+d)\ !}{a\ !\ b\ !\ c\ !} $

D) None of these

Show Answer

Answer:

Correct Answer: B

Solution:

  • Total number of books $ =a+2b+3c+d $ Since there are $ b $ copies of each of two books, $ c $ copies of each of three books and single copies of $ d $ books. Therefore the total number of arrangements is $ \frac{(a+2b+3c+d)\ !}{a\ !\ {{(b\ !)}^{2}}{{(c\ !)}^{3}}} $ .



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