Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-16 Discrete Mathematics — December 2015 sitting. 12 questions, 10 marks each (answer 10 of 12 per the paper; every question is solved here as a full study resource).
Reference texts: Rosen, Discrete Mathematics and Its Applications, 7th ed. (primary); Stewart, Calculus: Early Transcendentals, 9th ed. (for the calculus argument in Question 7).
Given. A standard 52-card deck (13 ranks × 4 suits), 5-card hands, sample space
size $\binom{52}{5}=2{,}598{,}960$; ranks run 2,3,…,10,J,Q,K,A (ace high only, no ace-low wrap).
Find. P(a) through P(e), each as a favorable-count / total-count ratio.
Approach. Count each event by choosing the rank pattern first (which ranks are used, and
with what multiplicity) and then independently choosing suits for each selected rank; divide every count by
$\binom{52}{5}$.
Given data
Quantity
Value
Deck size
52 cards (13 ranks × 4 suits)
Hand size
5 cards
Total hands $\binom{52}{5}$
2,598,960
(a) Five cards of consecutive rank. With ranks 2…A (13 ranks, ace high only), a
run of 5 consecutive ranks can start at 2,3,…,10 — that is $13-5+1=9$ possible rank windows.
Suits are unrestricted (any suit allowed to repeat across the 5 different ranks), so each of the 5 chosen
ranks independently picks 1 of 4 suits:
$$\#=9\times4^5=9\times1024=9{,}216 \qquad
P(a)=\boxed{\dfrac{9{,}216}{2{,}598{,}960}\approx0.003546}$$
(b) At least one card of every suit. With 5 cards and 4 suits, the pigeonhole principle
forces exactly one suit to appear twice and the other three suits once each. Choose which suit is doubled
($\binom41$), choose 2 ranks from that suit ($\binom{13}2$), and independently choose 1 rank from each of the
other three suits ($13^3$):
$$\#=\binom41\binom{13}2\,13^3=4\times78\times2197=685{,}464 \qquad
P(b)=\boxed{\dfrac{685{,}464}{2{,}598{,}960}\approx0.263745}$$
(c) All five cards the same suit (flush, straight flush included). Choose the suit
($\binom41$) and any 5 of its 13 ranks ($\binom{13}5$):
$$\#=\binom41\binom{13}5=4\times1287=5{,}148 \qquad
P(c)=\boxed{\dfrac{5{,}148}{2{,}598{,}960}\approx0.001981}$$
(d) Exactly one pair. Choose the paired rank ($\binom{13}1$), the 2 suits forming the
pair ($\binom42$), then 3 further ranks — all distinct from the pair rank and from each other
($\binom{12}3$) — each independently choosing 1 of 4 suits ($4^3$):
$$\#=\binom{13}1\binom42\binom{12}3\,4^3=13\times6\times220\times64=1{,}098{,}240 \qquad
P(d)=\boxed{\dfrac{1{,}098{,}240}{2{,}598{,}960}\approx0.422569}$$
(e) Full house. Choose the triple's rank ($\binom{13}1$) and its 3 suits out of 4
($\binom43$), then the pair's rank from the 12 remaining ranks ($\binom{12}1$) and its 2 suits ($\binom42$):
$$\#=\binom{13}1\binom43\binom{12}1\binom42=13\times4\times12\times6=3{,}744 \qquad
P(e)=\boxed{\dfrac{3{,}744}{2{,}598{,}960}\approx0.001441}$$
