Question

1. Sam and Janet are picking pets (dogs and/or cats) from among 14 different dogs and 11 different cats at the animal shelter.

How many different outcomes are there if Sam picks a dog and a cat and Janet picks either two dogs or two cats, but then Janet decides that she wants to swap exactly one animal with Sam, for his animal of the same kind (i.e. if Janet has picked cats, she swaps one of them with Sam's chosen cat)?

For example, suppose Sam picks the dog named "Astro" and the cat named "Bigglesworth", while Janet picks the dogs named "Chance" and "Danger"; then certainly Sam will get "Bigglesworth" and Janet will get "Astro", but there are two possible final situations: Sam will either get "Chance" or "Danger" in the end, and Janet will end up with the other.

(An "outcome" is a selection of animals for Sam and Janet.)

Answer 1

Solution

Back-up Theory

If an Activity 1 can be done in n ways, another Activity 2 can be done in m ways and for every

one way of doing Activity 1, there are m ways of doing Activity 2, then Activity 1 and Activity 2

can be simultaneously done in (n x m) ways. This is the Rule of Multiplication applicable to

both Permutations and Combinations..................................................................................................................................... (1)

Number of ways of selecting r things out of n things is given by ⁿC_r = (n!)/{(r!)(n - r)!}…….....................................................(2)

Values of ⁿC_r can be directly obtained using Excel Function: Math & Trig COMBIN…....................................................…. (2a)

Now to work out the solution,

Approach

Solution is obtained in two steps.

In Step 1, the given conditions are conceptualised into scenarios.

In Step 2, the above scenarios are analytically converted using the principles of Combinations enunciated under Back-up Theory to get the final count.

Step1:

Scenario 1

a) Sam picks 1 dog, say Di and 1 cat, say Cj

b) Janet picks 2 dogs, say Dk and Dn

c) Janet swaps Dk with Di

Scenario 2

a) and (b) same as in Scenario 1

c) Janet swaps Dn with Di

Scenario 3

a) same as in Scenario 1

b) Janet picks 2 cats, say Cm and Cp

c) Janet swaps Cm with Cj

Scenario 4

a) and (b) same as in Scenario 3

c) Janet swaps Cp with Di

Step 2:

Scenario 1

a) With 14 dogs and 11 cats, vide (2) and (1), number of selections = ¹⁴C₁ x ¹¹C₁ = 14 x 11 = 154. (3)

b) Since Sam has already picked 1 dog, now there are only 13 dogs. Vide (2) and (1), number of selections = ¹³C₂ = 78.... (4) c) There is only one possibility.................................................................................................................................................. (5)

Thus, vide (1), (3), (4) and (5), total count = 154 x 78 x 1 = 12012 ..........................................................................................(6)

Scenario 2

By the very analysis as in Scenario 1, total count = 154 x 78 x 1 = 12012 ..............................................................................(7)

Scenario 3

a) Same as in Scenario 1. Number of selections = 154......................................................................................................... (8)

b) Since Sam has already picked 1 cat, now there are only 10 cats. Vide (2) and (1), number of selections = ¹⁰C₂ = 45..... (9)

c) There is only one possibility............................................................................................................................................... (10)

Thus, vide (1), (8), (9) and (10), total count = 154 x 45 x 1 = 6930 .......................................................................................(11)

Scenario 4

By the very analysis as in Scenario 3, total count = 6930........................ ............................................................................(12)

Thus, the final count = (6) + (7) + (11) + (12) = 2(12012 + 6930)

= 37884 Answer

DONE