What is the permutation of 14 taken 5? 3. A group of 45 people is going to run a race. In how many possible ways can they be arranged as first, second, and third placers? 4. A group of 10 students wants to elect a president, vice president, secretary, and treasurer. How many different ways can they choose the officers? 5. How many ways can you introduce the 5 starting players of the Wildcats Basketball Team in the basketball game?
The Correct Answer and Explanation is:
Answers
- P(9,3)=504P(9,3)=504
- P(14,5)=240,240P(14,5)=240{,}240
- 85 140 possible podium orders
- 5 040 possible officer slates
- 120 possible introduction orders
Why those numbers are correct
A permutation counts the number of ways to arrange rr distinct items chosen from a pool of nn distinct items — with order mattering. The standard symbol is P(n,r)P(n,r) or nPrnP r. The counting rule comes straight from the Fundamental Principle of Counting (“multiply the number of choices you have at each step”).
Mathematically, P(n,r)=n!(n−r)!,P(n,r)=\frac{n!}{(n-r)!},
because n!n! lists every possible full ordering of nn objects, and dividing by (n−r)!(n-r)! cancels the trailing positions we no longer care about.
- P(9,3)P(9,3).
For the first slot you may choose any of 9 elements, then 8 remain, then 7 remain.
9×8×7=504.9\times 8\times 7=504. - P(14,5)P(14,5).
Multiply the first five descending integers from 14:
14×13×12×11×10=240 240.14\times13\times12\times11\times10=240\,240. - Race podium from 45 runners.
Gold, silver, and bronze are distinct, so order counts: 45P345P3.
45×44×43=85 14045\times44\times43=85\,140 possible podium line-ups. - Electing four different officers from 10 students.
Positions (president, vice president, secretary, treasurer) are inherently ordered, so again a permutation: 10P410P4.
10×9×8×7=5 04010\times9\times8\times7=5\,040.
(Notice that if we merely chose a four-person committee with no roles, we’d use combinations, but roles make order relevant.) - Introducing a five-player basketball line-up.
All five starters must be named, and the announcer controls the sequence. That is a complete ordering of five objects: 5P5=5!=1205P5=5!=120 possible sequences.
- P(9,3)P(9,3).
For the first slot you may choose any of 9 elements, then 8 remain, then 7 remain.
9×8×7=504.9\times 8\times 7=504. - P(14,5)P(14,5).
Multiply the first five descending integers from 14:
14×13×12×11×10=240 240.14\times13\times12\times11\times10=240\,240. - Race podium from 45 runners.
Gold, silver, and bronze are distinct, so order counts: 45P345P3.
45×44×43=85 14045\times44\times43=85\,140 possible podium line-ups. - Electing four different officers from 10 students.
Positions (president, vice president, secretary, treasurer) are inherently ordered, so again a permutation: 10P410P4.
10×9×8×7=5 04010\times9\times8\times7=5\,040.
(Notice that if we merely chose a four-person committee with no roles, we’d use combinations, but roles make order relevant.) - Introducing a five-player basketball line-up.
All five starters must be named, and the announcer controls the sequence. That is a complete ordering of five objects: 5P5=5!=1205P5=5!=120 possible sequences.
Key takeaway: whenever the wording distinguishes first/second/third, roles, or sequence in any way, you should expect a permutation, not a combination. The factorial shortcut nPr=n! / (n−r)!nP r=n!\!/\!(n-r)! provides a quick, reliable path: list the first rr factors of n!n! and multiply. The examples above simply plug specific nn and rr values into that universal rule.
