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Question: A library has ‘a’ copies of one book, ‘b’ copies each of the two books, ‘c’ copies each of three boo...

A library has ‘a’ copies of one book, ‘b’ copies each of the two books, ‘c’ copies each of three books, and a single copy of d books. The total number of ways in which these books can be arranged in a shelf is equal to
A. (a+2b+3c+d)!a!(b!)2(c!)3\dfrac{\left( a+2b+3c+d \right)!}{a!{{\left( b! \right)}^{2}}{{\left( c! \right)}^{3}}}
B. (a+2b+3c+d)!a!(2b!)(c!)3\dfrac{\left( a+2b+3c+d \right)!}{a!\left( 2b! \right){{\left( c! \right)}^{3}}}
C. (a+b+3c+d)!(c!)3\dfrac{\left( a+b+3c+d \right)!}{{{\left( c! \right)}^{3}}}
D. (a+2b+3c+d)!a!(2b!)(3c!)\dfrac{\left( a+2b+3c+d \right)!}{a!\left( 2b! \right)\left( 3c! \right)}

Explanation

Solution

In this problem, we have to find the total number of ways in which these books can be arranged in a shelf equal to which of the given options. We should know that, the number of ways of arranging n unlike objects in a line in n! for example, the different ways the a, b, c, d can be arranged. The number of ways of arranging n objects of which a of one type are alike, b of second type are alike and c of third type are alike is n!a!b!c!\dfrac{n!}{a!b!c!}, where n=a+b+c+dn=a+b+c+d.

Complete step by step solution:
We know that the given data is,
A library has ‘a’ copies of one book, ‘b’ copies each of the two books, ‘c’ copies each of three books, and a single copy of d books.
We can write this above statement in an equation as,
The total number of books, n = a+2b+3c+da+2b+3c+d .
The number of ways of arranging n objects of which a of one type are alike, b of second type are alike and c of third type are alike is n!a!b!c!\dfrac{n!}{a!b!c!}.
Now we can write the equation as,
(a+2b+c+d)!a!(b!)2(c!)3\Rightarrow \dfrac{\left( a+2b+c+d \right)!}{a!{{\left( b! \right)}^{2}}{{\left( c! \right)}^{3}}}
Therefore, the option A. (a+2b+3c+d)!a!(b!)2(c!)3\dfrac{\left( a+2b+3c+d \right)!}{a!{{\left( b! \right)}^{2}}{{\left( c! \right)}^{3}}} is the correct answer.

Note: We should always remember that the number of ways of arranging n objects of which a of one type are alike, b of second type are alike and c of third type are alike is n!a!b!c!\dfrac{n!}{a!b!c!} where,n=a+b+c+dn=a+b+c+d. We should also concentrate on the part of forming an equation for the total number of books from the given data.