Question
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!(b!)2(c!)3(a+2b+3c+d)!
B. a!(2b!)(c!)3(a+2b+3c+d)!
C. (c!)3(a+b+3c+d)!
D. a!(2b!)(3c!)(a+2b+3c+d)!
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 a!b!c!n!, where n=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+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 a!b!c!n!.
Now we can write the equation as,
⇒a!(b!)2(c!)3(a+2b+c+d)!
Therefore, the option A. a!(b!)2(c!)3(a+2b+3c+d)! 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 a!b!c!n! where,n=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.