Question Video: Solving Permutation Equations for Unknowns in the π‘Ÿ Position Mathematics

Solve the following equation for π‘Ÿ: 5π‘ƒπ‘Ÿ = 120.

03:17

Video Transcript

Solve the following equation for π‘Ÿ: five Pπ‘Ÿ is equal to 120.

We know that we calculate 𝑛Pπ‘Ÿ by taking 𝑛 factorial over 𝑛 minus π‘Ÿ factorial. In this case, we don’t know the π‘Ÿ-value. We know that our set has five elements, but we don’t know how many we’re trying to choose. To find π‘Ÿ, we need to find out how many decreasing consecutive integers, starting with five, we should multiply together to equal 120. We know that 𝑛 factorial is equal to 𝑛 times 𝑛 minus one factorial. Five Pπ‘Ÿ is then equal to five factorial over five minus π‘Ÿ factorial. If we expand the factorial in the numerator, we get five times four times three times two times one. And we know that five Pπ‘Ÿ must be equal to 120, but five factorial equals 120. So we end up with the equation 120 equals 120 over five minus π‘Ÿ factorial.

If we multiply both sides of the equation by five minus π‘Ÿ factorial, we get 120 times five minus π‘Ÿ factorial equals 120. And then if we divide both sides by 120, on the left, we have five minus π‘Ÿ factorial, and on the right, 120 divided by 120 equals one. This means we need an π‘Ÿ-value that will make five minus π‘Ÿ factorial equal to one. Based on the properties of factorials, we know that there are two places where a factorial equals one, zero factorial and one factorial, which means five minus π‘Ÿ must be zero or five minus π‘Ÿ must be one. If five minus π‘Ÿ is zero, then π‘Ÿ equals five. And if five minus π‘Ÿ is one, then π‘Ÿ equals four.

Remember, at the beginning, we said that π‘Ÿ would be equal to the number of decreasing consecutive integers beginning with five we multiply together to get 120. And so there is one other strategy we can use when we know the 𝑛-value of a permutation but we don’t know the π‘Ÿ-value. We know the number of permutations we can have is 120. And we know that we are beginning with a set of five. If we start by dividing 120 by five, we get 24. Now we take 24 and we divide by the consecutive integer below five. So 24 divided by four equals six.

Again, we’ll take that value and divide it by the integer that is below four. So six divided by three equals two. Two divided by the integer below three, which is two, equals one, which shows us that π‘Ÿ could equal four, that four consecutive decreasing integers beginning with five multiply together to equal 120. Two times three times four times five does equal 120. However, we could follow the pattern one final time because one divided by one equals one and one times two times three times four times five also equals 120, which gives us π‘Ÿ equals four or π‘Ÿ equals five.

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