Compute P(5,2). (Enter an exact number.) Compute C(7,5). (Enter an exact number.)
The Correct Answer and Explanation is:
Let’s compute each part:
1. Compute P(5,2)P(5, 2)P(5,2):
This is a permutation problem, which counts how many ways you can arrange 2 items out of 5 in a specific order.
The formula for permutations is:P(n,r)=n!(n−r)!P(n, r) = \frac{n!}{(n – r)!}P(n,r)=(n−r)!n!
So,P(5,2)=5!(5−2)!=5!3!P(5, 2) = \frac{5!}{(5 – 2)!} = \frac{5!}{3!}P(5,2)=(5−2)!5!=3!5!=5×4×3×2×13×2×1=1206=20= \frac{5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} = \frac{120}{6} = 20=3×2×15×4×3×2×1=6120=20
✅ Answer: P(5,2)=20P(5, 2) = 20P(5,2)=20
2. Compute C(7,5)C(7, 5)C(7,5):
This is a combination problem, which counts how many ways you can choose 5 items out of 7 without considering order.
The formula for combinations is:C(n,r)=n!r!(n−r)!C(n, r) = \frac{n!}{r!(n – r)!}C(n,r)=r!(n−r)!n!
So,C(7,5)=7!5!⋅(7−5)!=7!5!⋅2!C(7, 5) = \frac{7!}{5! \cdot (7 – 5)!} = \frac{7!}{5! \cdot 2!}C(7,5)=5!⋅(7−5)!7!=5!⋅2!7!=7×6×5!5!⋅2×1=42×5!5!⋅2=422=21= \frac{7 \times 6 \times 5!}{5! \cdot 2 \times 1} = \frac{42 \times 5!}{5! \cdot 2} = \frac{42}{2} = 21=5!⋅2×17×6×5!=5!⋅242×5!=242=21
✅ Answer: C(7,5)=21C(7, 5) = 21C(7,5)=21
Explanation (Approx. 300 Words)
In combinatorics, permutations and combinations are used to count how many ways objects can be arranged or selected.
A permutation refers to an arrangement of objects where order matters. The formula for the number of permutations of rrr objects from a set of nnn is:P(n,r)=n!(n−r)!P(n, r) = \frac{n!}{(n – r)!}P(n,r)=(n−r)!n!
In our case, P(5,2)P(5, 2)P(5,2) asks how many ways we can choose 2 objects from a group of 5 and arrange them in sequence. We calculate 5!5!5! (which is 120), and divide it by 3!3!3! (which is 6), giving 20 distinct ordered arrangements.
On the other hand, a combination refers to a selection of objects where order does not matter. The formula is:C(n,r)=n!r!(n−r)!C(n, r) = \frac{n!}{r!(n – r)!}C(n,r)=r!(n−r)!n!
For C(7,5)C(7, 5)C(7,5), we calculate how many ways we can choose 5 items from 7 where order is irrelevant. The value comes out as 21, meaning there are 21 different ways to choose 5 items from 7 without considering their order.
Understanding the difference between permutations and combinations is important in probability, statistics, and many real-world applications like lottery games, password generation, and arranging teams.
