Daily · 2026-05-09

Three problems · Combinatorics, Compound Interest, Algebra

One bite-sized math problem set for the day. Read the statement, think it through, then expand the solution to check your reasoning.

#1 · Combinatorics·Medium

From a group of 10 people, how many ways can a committee of 3 be chosen?

A. $30$
B. $120$
C. $720$
D. $1000$

Solution

\binom{10}{3} = \dfrac{10!}{3! \cdot 7!} = \dfrac{10 \cdot 9 \cdot 8}{3!} = \dfrac{720}{6} = 120

.

The convention: 'committee' implies order doesn't matter — a committee of {Alice, Bob, Carol} is the same as {Carol, Alice, Bob}. That's exactly when you use

\binom{n}{k}

instead of

P(n,k)

Combinations vs. permutations is one of the highest-leverage distinctions in counting. Whenever you read a problem, ask first: does the order matter? 'Lineup' / 'rank' / 'arrangement' = permutations. 'Group' / 'committee' / 'team' = combinations.

#2 · Compound Interest·Hard

You invest \$200 at the end of each month into an account paying 6% APR, compounded monthly. After 30 years, approximately how much is in the account?

A.\$72,000
B.\$120,000
C.\$200,000
D.\$1,000,000

Solution

This is a future-value-of-annuity problem:

FV = P \cdot \frac{(1+i)^n - 1}{i}

where

P = 200

i = 0.06/12 = 0.005

n = 30 \cdot 12 = 360

(1.005)^{360} \approx 6.023

. So

FV \approx 200 \cdot \dfrac{6.023 - 1}{0.005} = 200 \cdot 1004.5 \approx \

200{,}900

.Note: you contributed only

200 \cdot 360 = \

72{,}000

. The other ~\$129{,}000 is compounding.

The single most important formula in personal finance. \

200/month for 30 years at 6\% turns into \

200K — almost tripling your contribution. The same calculation at 8\% gives ~\

300K; at 10\%, ~\

450K. Small differences in rate, huge differences in outcome — that's the geometric-series doing its job.

#3 · Algebra·Easy

f(x) = x^2 + 1

and

g(x) = 2x - 3

, what is

f(g(2))

A. $1$
B. $2$
C. $5$
D. $11$

Solution

Inside out. First:

g(2) = 2 \cdot 2 - 3 = 1

. Then:

f(1) = 1^2 + 1 = 2

Function composition: evaluate the innermost first. Easy to mess up if you confuse

f(g(x))

with

g(f(x))

— they generally differ. Here

g(f(2)) = g(5) = 7 \ne 2

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