Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examination, 04-BS-16 Discrete Mathematics, undated sitting (May 2019). Closed book; approved Casio or Sharp calculator only. The exam instructs "answer 10 of the 12 questions"; every question is answered below as a complete study resource.
Source note: This paper is the May 2019 sitting (every page footer reads "04-BS-16/May 2019"). Two printed statements are defective as set and are flagged where they occur: Question 7(a) prints the last term of $\{1,5,9,\dots\}$ as $4n-1$ (the pattern and the stated sum require $4n-3$), and Question 8(b) prints "$n>2$" although $4^n>n^4$ fails at $n=3,4$. Question 12(c)'s parameters also make a connected graph impossible; this is noted at that part.
Reference texts: Rosen, Discrete Mathematics and Its Applications, 7th ed. (Pearson) — used throughout for logic, set theory, functions, combinatorics, probability, induction, asymptotic (Big-O) notation, and graph theory.
Given. ENGINEERING has 11 letters: E×3, N×3, G×2, I×2, R×1.
Find. (a) total distinct arrangements. (b)–(e) arrangements under each stated constraint.
Approach. Use the multiset-permutation formula $n!/(n_1!n_2!\cdots)$ throughout, fixing letters as required by each constraint and counting arrangements of what remains.
a) Total distinct permutations. $11$ letters with repeats E:3, N:3, G:2, I:2, R:1: $$\frac{11!}{3!\,3!\,2!\,2!\,1!}=\frac{39916800}{6\cdot6\cdot2\cdot2\cdot1}=\frac{39916800}{144}=\boxed{277200}.$$
b) Start with R, end with G. Fixing the only R first and one G last uses up those two letters, leaving 9 letters E:3, N:3, G:1, I:2 to arrange freely in the middle: $$\frac{9!}{3!\,3!\,1!\,2!}=\frac{362880}{6\cdot6\cdot1\cdot2}=\frac{362880}{72}=\boxed{5040}.$$
c) Start with a vowel. The vowels present are E and I. Count arrangements starting with E, plus arrangements starting with I, separately (mutually exclusive). Starting with E fixes one E, leaving 10 letters E:2,N:3,G:2,I:2,R:1: $10!/(2!3!2!2!1!)=3628800/48=75600$. Starting with I fixes one I, leaving 10 letters E:3,N:3,G:2,I:1,R:1: $10!/(3!3!2!1!1!)=3628800/72=50400$. Sum: $\boxed{75600+50400=126000}$.
d) All three E's together (as the block "EEE"). Glue the 3 E's into one unit, leaving 9 units to arrange: {EEE-block, N,N,N, G,G, I,I, R} $=9$ units with N:3, G:2, I:2, block:1, R:1: $$\frac{9!}{3!\,2!\,2!\,1!\,1!}=\frac{362880}{6\cdot2\cdot2}=\frac{362880}{24}=\boxed{15120}.$$
e) Start with ING and end with ING. This consumes I:2, N:2, G:2 (both occurrences of each) in the two fixed 3-letter blocks, leaving $11-6=5$ middle letters: E:3, N:$3-2=1$, R:1. Arranging the 5 middle letters: $$\frac{5!}{3!\,1!\,1!}=\frac{120}{6}=\boxed{20}.$$
