Calculating combinations Practice Questions (Grade 11)

Warm-upeasy

There are $4$ posters, and $3$ will be chosen for the hallway. What is $C(4, 3)\textsf{?}$

$C(4,3) = \dfrac{4!}{3! \cdot (4-3)!} = \dfrac{24}{6 \cdot 1} = 4\textsf{.}$

Mid-gameeasy

A pizza shop offers $8$ toppings, and a pizza gets $3$ of them. What is $C(8, 3)\textsf{?}$

$C(8,3) = \dfrac{8!}{3! \cdot 5!} = \dfrac{40320}{6 \cdot 120} = 56\textsf{.}$

Mid-gameeasy

A class has $6$ students, and $3$ will be chosen for a group. What is $C(6, 3)\textsf{?}$

$C(6,3) = \dfrac{6!}{3! \cdot 3!} = \dfrac{720}{6 \cdot 6} = 20\textsf{.}$

Mid-gameeasy

An art project has $7$ colors, and $4$ will be chosen for the poster. What is $C(7, 4)\textsf{?}$

$C(7,4) = \dfrac{7!}{4! \cdot 3!} = \dfrac{5040}{24 \cdot 6} = 35\textsf{.}$

Mid-gameeasy

A yearbook has $6$ photos, and $1$ will be chosen for the cover. What is $C(6, 1)\textsf{?}$

$C(6,1) = \dfrac{6!}{1! \cdot 5!} = \dfrac{720}{1 \cdot 120} = 6\textsf{.}$

Mid-gameeasy

A team has $5$ players, and all $5$ will be chosen for the photo. What is $C(5, 5)\textsf{?}$

$C(5,5) = \dfrac{5!}{5! \cdot 0!} = \dfrac{120}{120 \cdot 1} = 1\textsf{.}$

Mid-gameeasy

A bakery has $8$ desserts, and $4$ will be chosen for a tasting plate. What is $C(8, 4)\textsf{?}$

$C(8,4) = \dfrac{8!}{4! \cdot 4!} = \dfrac{40320}{24 \cdot 24} = 70\textsf{.}$

Buzzer beatereasy

A fair lists $8$ booths, and a student will visit $2$ of them. What is $C(8, 2)\textsf{?}$

$C(8,2) = \dfrac{8!}{2! \cdot 6!} = \dfrac{40320}{2 \cdot 720} = 28\textsf{.}$

Calculating combinations (ACT). Game on.