Question

How many distinguishable arrangements can be made from three different algebra books and four different geometry books?

a) without restrictions?

b)if the algebra books must appear to the left of the geometry books?

c)if the books on a particular subject must appear together

Answer 1

Answer:

There are 3 algebra books and 4 geometry books. So total books are 7.

So there are 7 positions and 7 books have to be arranged.

a)

We know that the number of ways of arranging n unlike objects in a line is n!

If no restriction , then 7 books can be arranged in 7! ways.

=7*6*5*4*3*2*1 = 5040 arrangemets.

b)

Now if the algebra books must appear to the left of the geometry books, then first algebra books can be arranged in 3! ways and geometry books can be arranged in 4! ways.

Total number of arrangemet are 3!*4!

= 3*2*1 * 4*3*2*1 = 144 arrangemets

c)

Now if the books on a particular subject must appear together.

Let us consider 1 set algebra books appearing together, and 1 set geometry books appearing together.

These two sets can be arranged in 2! ways. In set of algebra books can be arranged in 3! ways and in set of algebra books can be arranged in 3!.

So total number of arrangements: = 2!*3!*4!

= 2 *1 (3*2*1) * (4*3*2*1) = 288 arrangemets