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Question Video: Determining Equivalent Expressions Involving Combinations and Permutations Mathematics

Which of the following is equal to (𝑛𝐢₁ Γ— 6𝑛𝐢₆)/7𝑛𝐢₇? [A] 6𝑛𝑃₆/(7𝑛 βˆ’ 7)𝑃₇ [B] 6𝑛𝑃₆/(7𝑛 βˆ’ 1)𝑃₆ [C] 6𝑛𝑃₆/(7𝑛 βˆ’ 7)𝑃₆ [D] 6𝑛𝑃₆/(7𝑛 βˆ’ 1)𝑃₇

03:17

Video Transcript

Which of the following is equal to 𝑛𝐢 one times six 𝑛𝐢 six over seven 𝑛𝐢 seven? (A) Six 𝑛𝑃 six over seven 𝑛 minus seven 𝑃 seven. (B) Six 𝑛𝑃 six over seven 𝑛 minus one 𝑃 six. (C) Six 𝑛𝑃 six over seven 𝑛 minus seven 𝑃 six. (D) Six 𝑛𝑃 six over seven 𝑛 minus one 𝑃 seven.

Almost immediately, we notice that our initial expression is in terms of combinations, three different combinations, while all of our answer choices are given as an expression in terms of permutations. So we need to rewrite these combinations in terms of permutations. But first, we notice this 𝑛𝐢 one, and the properties of combinations tell us that 𝑛𝐢 one just equals 𝑛. To choose one item from a set of 𝑛, there are 𝑛 different ways to do that. So in our first step, we’ll substitute 𝑛 in place of 𝑛𝐢 one.

In our next step, we’ll want to rewrite these combinations as permutations. We can do that by recognizing that π‘›πΆπ‘Ÿ equals π‘›π‘ƒπ‘Ÿ over π‘Ÿ factorial, which means we have 𝑛 times six 𝑛𝑃 six over six factorial divided by seven 𝑛𝑃 seven over seven factorial. And then in place of that division, we’ll multiply by the reciprocal such that we have 𝑛 times six 𝑛𝑃 six over six factorial times seven factorial over seven 𝑛𝑃 seven. We have six factorial in the denominator and seven factorial in the numerator. Recall that we can rewrite seven factorial as seven times six factorial. This means seven factorial over six factorial simplifies to seven. From there, we have seven 𝑛 times six 𝑛𝑃 six over seven 𝑛𝑃 seven.

Notice that none of the four answer choices have a seven 𝑛 term, which means we need to think of some way to simplify this further. To do that, we notice that we have a seven 𝑛 term and an 𝑛 equals seven 𝑛. A property of permutations tells us that π‘›π‘ƒπ‘Ÿ is equal to 𝑛 times 𝑛 minus one π‘ƒπ‘Ÿ minus one. That means seven 𝑛𝑃 seven equals seven 𝑛 times seven 𝑛 minus one 𝑃 seven minus one. And in place of that seven minus one, we’ll have six. In our denominator, in place of seven 𝑛𝑃 seven, we can substitute the term seven 𝑛 times seven 𝑛 minus one 𝑃 six. And then the seven 𝑛 in the numerator and the denominator cancel, which gives us a simplified form of six 𝑛𝑃 six over seven 𝑛 minus one 𝑃 six, which is option (B).

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