Question

Principles of Counting:

You have a ten volume set of books.

a) How many different ways can the books be arranged?

b) How many of these arrangements have Volume 2 somewhere to the right of Volume 1 and Volume 3 somewhere to the right of Volume 2.

c) You move the books to a shelf that only holds 7 of your books. How many ways are there to arrange the seven books on this shelf? Further, how many of those ways have the smallest-numberered volume on the shelf to the far left? (For example, 2, 9, 10, 7, 3, 5, 8)

Answer #1

Solution:

If we want to arrange 10 books so different no. Of ways they can be arranged in shelf = 10!

= 10*9*8*7*6*5*4*3*2*1 = 3628800

Solution(b)

3,2,1 is one combination and 7 books can be arranged anywhere

So total no. Of ways they can be arranged = 8!

=8×7×6×5×4×3×2×1 = 40320

Solution(c)

If we need to select 7 books out of 10 books than no. Of ways we can arrange = 10C7 =(10×9×8)/(3×2) = 120 ways

Now if smallest no. Is 1 than we need to select 2 more book out of 9. So total no. Of ways = (9C2) = (9×8)/2 = 36

If smallest no. Is 2 than we need to select 2 more book out of 8 sp total no. Of ways = (8C2) = (8×7)/2 = 28

If smallest no. Is 3 than we need to select 2 more book out of 7 so total no .of ways = 7C2 =(7*6)/2 =21

If smallest no. Is 4 than we need to select 2 more books out of 6 books so total no of ways =(6×5)/2= 15

So total no .of ways = 36+28+21+15 = 100 ways

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