Question

I have nine different programming textbooks on my bookshelf, five C++ and four Java. In how many ways can I arrange the books a) if there are no restrictions? b) ifthe languages should alternate? c) if all the C++books must be next to each other? d) if all the C++books must be next to each other and all the Java books must be next to each other? 1. 2. Suppose that you draw five cards from a standard deck of 52. a) How many ways can you draw exactly 2 spades? b) How many ways can you draw at most 2 spades? c) How many ways can you draw 3 spades and 2 hearts? d) How many ways can you draw exactly 1 spade, 2 hearts and 2 kings?

Answer 1

Solution

Q1

Back-up Theory

Number of ways of arranging n distinct things among themselves (i.e.,, permutations)

= n!

= n(n - 1)(n - 2) …… 3.2.1..............................................................................................………………………………….….(1)

Values of n!can be directly obtained using Excel Function: Math & Trig FACT (Number)................................ (1a)

Number of ways of arranging n distinct things among themselves such that p out of n things are always together

= (n – p + 1)! x p! ………..………………………………………………………………….….(2)

Number of ways of arranging n distinct things among themselves such that p out of n things are always together and also (n – p) are always together

= 2!(n – p)! x p! ………..………………………………………………………………….….(3)

Now, to work out the solution,

Here

n = 9, ................................................................................................................................... (4a)

p = 5 (C++) ................................................................................................................................... (4b)

(n - p) = 4 (Java) ................................................................................................................................... (4c)

Part (a)

Vide (4a) and (1), number of arrangements = 9! = 362880 [vide (1a)] Answer 1

Part (b)

The arrangement needs to be: CJCJCJCJC only.

Now, C can occupy 5 odd positions in 5! ways and for each of these arrangements, J can occupy 4 even positions in 4! ways. Thus, total number of arrangements = 5! X 4! = 120 x 24 = 2880 Answer 2

Part (c)

Vide (4a), (4b) and (2), number of arrangements = 5! x 5! = 14400 Answer 3

Part (d)

Vide (4a), (4b), (4c) and (3), number of arrangements = 2! x 5! x 4! = 5760 Answer 4

DONE