Problem

Set 2.8
Statistics & Probability COMBINATIONS

Analyze the given statements. If the statement presents a Combination problem, solve for
the given problems accordingly, however, if the statement presents a Permutation
problem, simply state that it is one and do not solve.

1) 8 kids join a race organized by the school. How many possible ways can the
organizers choose a 1st placer, a 2nd placer, and a 3rd placer?

Order is important here due to the fact that the problem asks for specific positions
(1st, 2nd, and 3rd placers) – this is then a PERMUTATION scenario (no need to solve).

2) 8 kids join a race organized by the school. How many possible ways can the
organizers choose three different winners?

Order is not important here as this simply asks for winners instead of placers
(imagine this – A winning 1st place, B winning 2nd place, and C winning 3rd place and
B winning 1st place, C winning 2nd place, and A winning 3rd place – these are two
different arrangements in awards (therefore a permutation), but the people
involved are unchanged (which is considered a combination) which implies that this
is a COMBINATION scenario.

Thus:

8C3 = 56

There are 56 different possible compositions of winners for this scenario.

3) There are 5 different positions for homeroom officers – the President, Vice-
President, Secretary, Treasurer, and Beadle. In a class of 30 students, how many
possible ways can you find a set of homeroom officers?

This is a COMBINATION scenario because it asks for a set of homeroom officers, not
so much of a concern of the individual positions.

Thus:

30C5 = 142, 506

There are 142, 506 different possible sets of homeroom officers in a class of 30
students.



4) A hand is a set of 5 cards drawn from a deck of playing cards. There are 52 total
cards in a deck of playing cards. There are 4 suits – hearts (red), diamonds (red),
clovers (black), and spades (black). For each suit, there are 13 cards – 3 of which are
face cards (King, Queen, and Jack), 10 of which are numerical cards (Ace up to 10).

How many hands are possible to draw from a standard deck of playing cards?

All cards are allowed to be used.

Thus:

52C5 = 2, 598, 960

There are 2, 598, 960 possible hands that can be drawn from a standard deck of
playing cards.

5) How many hands are possible to draw from a standard deck of playing cards such
that there are exactly 2 Kings in the hand?

Such problems may simply be answered using slots. In a hand, there are 5 slots, as
such:

∙ ∙ ∙ ∙ ∙

What one is sure of is that two of the slots are exactly 2 Kings. Imagining that the
first two slots are the Kings, they can then be represented as 4C2 (because there are
4 total Kings in the deck of cards, and 2 of them will be chosen).

The 2 remaining Kings will not be included back in the deck because the hand
specifically states that there should only be 2 Kings – this means that there are only
48 cards left to be used for the remaining 3 slots. This is then represented with
48C3.

Thus:

4C2 ∙ 48C3 = 103, 776

There are 103, 776 possible hands that have exactly 2 Kings.






6) How many hands are possible to draw from a standard deck of playing cards such
that there are at least 2 red face cards and 2 black numerical cards?

Such problems may simply be answered using slots. In a hand, there are 5 slots, as
such:

∙ ∙ ∙ ∙ ∙

What one is sure of is that two of the slots are 2 red face cards. Imagining that the
first two slots are the red face cards, they can then be represented as 6C2 (because
there are 3 face cards per suit, and there are 2 suits considered as red).

What one is sure of is that two of the slots are 2 black numerical cards. Imagining
that the next two slots are the black numerical cards, they can then be represented
as 20C2 (because there are 10 numerical cards per suit, and there are 2 suits
considered as black).

The last slot is simply for the remaining card which is still 48 because only 4 cards
have been used up so far for the previous conditions (the unused cards from the
previous conditions are still placed in the deck to be drawn from because the
problem uses the phrase at least instead of exactly – this means that it is alright if
there will be more than 2 red face cards in the hand or more than 2 black numerical
cards in the hand).

Thus:

6C2 ∙ 20C2 ∙ 48C1 = 136, 800

There are 136, 800 possible hands that have at least 2 red face cards and 2 black
numerical cards.