Calculating permutations Practice Questions (Grade 11)

Warm-upeasy

There are $4$ posters, and $3$ will be hung left to right. What is $P(4, 3)\textsf{?}$

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

Mid-gameeasy

A pizza shop offers $8$ toppings, and a pizza ranks $3$ of them from first to third. What is $P(8, 3)\textsf{?}$

$P(8,3) = \dfrac{8!}{5!} = \dfrac{40320}{120} = 336\textsf{.}$

Mid-gameeasy

A class has $6$ students, and $3$ will speak first, second, and third. What is $P(6, 3)\textsf{?}$

$P(6,3) = \dfrac{6!}{3!} = \dfrac{720}{6} = 120\textsf{.}$

Mid-gameeasy

An art project has $7$ colors, and $4$ will be painted in order on a striped poster. What is $P(7, 4)\textsf{?}$

$P(7,4) = \dfrac{7!}{3!} = \dfrac{5040}{6} = 840\textsf{.}$

Mid-gameeasy

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

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

Mid-gameeasy

A team has $5$ players, and all $5$ will be written in a batting order. What is $P(5, 5)\textsf{?}$

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

Mid-gameeasy

A bakery has $8$ desserts, and $4$ will be tasted in order. What is $P(8, 4)\textsf{?}$

$P(8,4) = \dfrac{8!}{4!} = \dfrac{40320}{24} = 1680\textsf{.}$

Buzzer beatereasy

A fair lists $8$ booths, and a student will visit a first booth and then a second booth. What is $P(8, 2)\textsf{?}$

$P(8,2) = \dfrac{8!}{6!} = \dfrac{40320}{720} = 56\textsf{.}$

Calculating permutations (ACT). Game on.