Basic Counting Principles: Additional Practice (Set 1)

Three positions in sequence, with $26$ choices per letter and $10$ for the digit, so multiply.

$26 \times 26 \times 10 = 6760$

Count each type by multiplying: a $2$-letter code gives $26 \times 26 = 676$ and a $3$-digit code gives $10 \times 10 \times 10 = 1000$. A ticket is one type or the other, so add.

$676 + 1000 = 1676$

Each flip has $2$ outcomes, repeated for all eight flips, so multiply eight twos.

$2^8 = 256$

Choose the digits in sequence: first digit $1$ through $9$ ($9$), the two middle digits any digit ($10$ each), and the last even ($5$).

$9 \times 10 \times 10 \times 5 = 4500$

One prize is chosen from three separate groups, so add the three group sizes.

$7 + 5 + 4 = 16$

Multiplying would count one item from each group, but only one prize is won.

Each die has $6$ outcomes, repeated for all three dice, so multiply three sixes.

$6 \times 6 \times 6 = 6^3 = 216$

Three independent choices in sequence, so multiply the three counts.

$4 \times 3 \times 5 = 60$

Choose the letters in sequence with no repeats, so the choices shrink from $26$.

$26 \times 25 \times 24 = 15600$

One representative is chosen from two separate groups (Class A or Class B), so add.

$12 + 15 = 27$

Each spin has $5$ outcomes, repeated for all three spins, so multiply three fives.

$5 \times 5 \times 5 = 5^3 = 125$

Four positions in sequence: $26$ choices for the letter and $10$ for each digit, so multiply.

$26 \times 10 \times 10 \times 10 = 26000$

The front spot is fixed for the youngest, so only the remaining four positions are free. Fill those four with the other four people, with the choices shrinking.

$4 \times 3 \times 2 \times 1 = 24$