Question Video: Using the Addition Rule of Counting Principles Mathematics

Michael is buying some stationery supplies for his office. He needs to buy 6 items, choosing from 20 types of pens, 10 types of pencils, and 5 types of printing paper. He must have at least 3 pens and only one paper package. Which of the following calculations represents the number of options Michael has when buying supplies? [A] ₂₀𝐢₃ + ₁₀𝐢₂ + 5 + ₂₀𝐢₄ + 10 + 5 + ₂₀𝐢₅ + 5 [B] ₂₀𝐢₃ Γ— ₁₀𝐢₂ Γ— 5 + ₂₀𝐢₄ Γ— 10 Γ— 5 + ₂₀𝐢₅ Γ— 5 [C] ₂₀𝐢₃ + ₁₀𝐢₂ + 5 + ₂₀𝐢₄ + 10 + 5 + ₂₀𝐢₅ [D] ₂₀𝐢₃ Γ— ₁₀𝐢₂ + ₂₀𝐢₄ Γ— 10 + ₂₀𝐢₅ [E] ₂₀𝐢₃ Γ— ₁₀𝐢₂ Γ— 5 + ₂₀𝐢₄ Γ— 10 Γ— 5 + ₂₀𝐢₅

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Video Transcript

Michael is buying some stationery supplies for his office. He needs to buy six items, choosing from 20 types of pens, 10 types of pencils, and five types of printing paper. He must have at least three pens and only one paper package. Which of the following calculations represents the number of options Michael has when buying supplies? Is it (A) 20 choose three plus 10 choose two plus five plus 20 choose four plus 10 plus five plus 20 choose five plus five. Option (B) 20 choose three multiplied by 10 choose two multiplied by five plus 20 choose four multiplied by 10 multiplied by five plus 20 choose five multiplied by five. Option (C) 20 choose three plus 10 choose two plus five plus 20 choose four plus 10 plus five plus 20 choose five. Option (D) 20 choose three multiplied by 10 choose two plus 20 choose four multiplied by 10 plus 20 choose five. Or option (E) 20 choose three multiplied by 10 choose two multiplied by five plus 20 choose four multiplied by 10 multiplied by five plus 20 choose five.

Let’s begin by considering the different ways Michael can select six items. We are told he must have at least three pens and only one paper package. One way of selecting the six items is to have three pens, two pencils, and one paper package. Alternatively, Michael could buy four pens, one pencil, and one pack of paper. His final option would be to buy five pens, zero pencils, and one pack of printing paper. There are three different ways that Michael could select the six items using the given constraints. In this question, the order that Michael selects the items doesn’t matter. And we know that the number of ways of choosing π‘Ÿ items from 𝑛 items when order doesn’t matter is 𝑛 choose π‘Ÿ.

We are told that there are 20 types of pens. This means that the number of ways of choosing three of these is equal to 20 choose three. There are 10 types of pencils, so the number of ways of choosing two of these is 10 choose two. As there are five types of printing paper, the number of ways of selecting one of these is equal to five choose one. We can repeat this for four pens, one pencil, and one paper package. We have 20 choose four, 10 choose one, and five choose one. Finally, when choosing five pens, zero pencils, and one paper package, we have 20 choose five, 10 choose zero, and five choose one.

As selecting pens, pencils, and paper are independent events, we can calculate the total number of ways of selecting three pens, two pencils, and one paper package by multiplying the number of ways of selecting three pens by the number of ways of selecting two pencils by the number of ways of selecting one paper package. We can repeat this process for selecting four pens, one pencil, and one paper package together with five pens, zero pencils, and one paper package. At this stage, it is worth recalling that 𝑛 choose zero is equal to one. This means that 10 choose zero is equal to one. We also know that 𝑛 choose one is equal to 𝑛. This means that 10 choose one equals 10 and five choose one is equal to five.

The number of ways of selecting three pens, two pencils, and one paper package is equal to 20 choose three multiplied by 10 choose two multiplied by five. Selecting four pens, one pencil, and one paper package is equal to 20 choose four multiplied by 10 multiplied by five. And choosing five pens and one paper package is equal to 20 choose five multiplied by one multiplied by five, which is the same as 20 choose five multiplied by five.

Next, we recall that if 𝐴 and 𝐡 are mutually exclusive events, where 𝐴 has π‘š distinct outcomes and 𝐡 has 𝑛 distinct outcomes, the total number of outcomes is π‘š plus 𝑛. This is known as the addition rule. And this means that the total number of options that Michael will have is the sum of the three products. This is equal to 20 choose three multiplied by 10 choose two multiplied by five plus 20 choose four multiplied by 10 multiplied by five plus 20 choose five multiplied by five. This corresponds to option (B), which is the number of ways Michael can buy six items including at least three pens and only one paper package.

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