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Home » Solution: We are given $ p + q = 5 $ and $ pq = 6 $. We use the identity: - ClickBalance

Solution: We are given $ p + q = 5 $ and $ pq = 6 $. We use the identity: - ClickBalance

📅 May 18, 2026 👤 scraface

May 18, 2026

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📌 A science administrator is organizing a research symposium with 5 different sessions, each requiring a unique topic from a pool of 10 available topics. In how many ways can the topics be assigned to the sessions?

📌 To solve this problem, we need to determine how many ways we can assign 5 unique topics to 5 sessions from a pool of 10 available topics. This is a problem of permutations since the order in which topics are assigned to sessions matters.

📌 The number of ways to choose 5 topics from 10 is given by the combination formula:

📌 Assign the selected topics to the sessions:

📌 Multiply the number of ways to choose the topics by the number of ways to assign them:

📌 A university professor is assigning 6 different research projects to 3 students such that each student receives exactly 2 projects. In how many distinct ways can the projects be distributed?

📌 Here, we must distribute 6 unique projects among 3 students, with each receiving exactly 2 projects. This can be approached through combinations and division by symmetry since the order in which students receive the projects doesnt matter.

📌 Divide the 6 projects into 3 groups of 2:

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rac{3(\sqrt{7} + \sqrt{2})}{5}
oxed{rac{3(\sqrt{7} + \sqrt{2})}{5}}
Question: Expand the product $(2x - 3y)(x^2 + 4xy - 5y^2)$.
We apply the distributive property to expand the product:
2x(x^2 + 4xy - 5y^2) = 2x^3 + 8x^2y - 10xy^2
-3y(x^2 + 4xy - 5y^2) = -3x^2y - 12xy^2 + 15y^3
$2x^3$
$8x^2y - 3x^2y = 5x^2y$
$-10xy^2 - 12xy^2 = -22xy^2$
$15y^3$

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