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. 6 men, 4 women (10 people total, including the three named individuals John, Mary and Susan
for part (b)); the word SCIENCE (7 letters: S,C,I,E,N,C,E, with C and E each repeated twice).
Find. The five counts (a)–(e) described above.
Approach. (a) treat the 4 women as one block and apply the circular-permutation formula
$(n-1)!$ to the resulting units, then multiply by the internal arrangements of the block. (b)/(c) split into
disjoint cases by how many special people / women are on the committee and sum $\binom{\cdot}{\cdot}$ products.
(d)/(e) use the multiset-permutation formula $n!/\prod(\text{repeat counts}!)$, treating tied letters as one
block for (e).
(a) Circular table, women block together. Treat the 4 women as a single unit; together
with the 6 men this gives $6+1=7$ units to seat around a circle. Since clockwise and counter-clockwise are
counted as different arrangements, the standard circular-permutation count $(7-1)!$ applies directly (no
extra division by 2 for reflection symmetry):
$$(7-1)! = 720 \text{ arrangements of the 7 units.}$$
The 4 women can be internally ordered within their block in $4!=24$ ways, independent of the block's position:
$$\#=720\times24=\boxed{17{,}280}$$
(b) Committee of 5, at most one of John/Mary/Susan. Split into two disjoint cases from
the 10 people (3 special + 7 ordinary):
0 special people: choose all 5 from the 7 non-special people: $\binom75=21$.
Exactly 1 special: choose 1 of the 3 special people and 4 of the 7 others: $\binom31\binom74=3\times35=105$.
$$\#=21+105=\boxed{126}$$
(c) Committee of 5, more women than men (4 women, 6 men available). With 5 seats and only
4 women available, "more women than men" restricts to (3W,2M) or (4W,1M) — (5W,0M) is impossible since
only 4 women exist:
$$\binom43\binom62 + \binom44\binom61 = (4\times15)+(1\times6)=60+6=\boxed{66}$$
(d) SCIENCE, no restrictions. SCIENCE has 7 letters with C repeated twice and E repeated
twice (S,I,N each appear once):
$$\#=\dfrac{7!}{2!\,2!}=\dfrac{5040}{4}=\boxed{1{,}260}$$
(e) SCIENCE, the two C's together. Glue the CC pair into one block, leaving 6 units to
arrange: [CC], S, I, E, N, E — E still repeats twice, no other repeats:
$$\#=\dfrac{6!}{2!}=\dfrac{720}{2}=\boxed{360}$$
