Solving Permutation Equation 9P5 + 5. 9P4 = 10 Pr

The problem asks us to find the value of \(r\) in the equation \(9P_5 + 5 \cdot 9P_4 = 10 P_r\). This involves understanding and applying the concept of permutations.

Recall the formula for permutations, \(nP_r\):

\(nP_r = \frac{n!}{(n-r)!}\)

Let's calculate the values of the permutations on the left side of the equation: \(9P_5\) and \(9P_4\).

Calculating 9P5

Using the permutation formula for \(9P_5\) with \(n=9\) and \(r=5\):

\(9P_5 = \frac{9!}{(9-5)!} = \frac{9!}{4!}\)

Expanding the factorials:

\(9P_5 = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1}\)

We can cancel out the \(4!\) terms:

\(9P_5 = 9 \times 8 \times 7 \times 6 \times 5\)

Calculating the product:

\(9P_5 = 72 \times 42 \times 5 = 72 \times 210 = 15120\)

Calculating 9P4

Using the permutation formula for \(9P_4\) with \(n=9\) and \(r=4\):

\(9P_4 = \frac{9!}{(9-4)!} = \frac{9!}{5!}\)

Expanding the factorials:

\(9P_4 = \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{5 \times 4 \times 3 \times 2 \times 1}\)

We can cancel out the \(5!\) terms:

\(9P_4 = 9 \times 8 \times 7 \times 6\)

Calculating the product:

\(9P_4 = 72 \times 42 = 3024\)

Substituting Values into the Equation

Now substitute the calculated values back into the original equation \(9P_5 + 5 \cdot 9P_4 = 10 P_r\):

\(15120 + 5 \times 3024 = 10 P_r\)

Calculate the left side:

\(15120 + 15120 = 10 P_r\)

\(30240 = 10 P_r\)

Solving for r in 10 Pr = 30240

We have the equation \(10 P_r = 30240\). Using the permutation formula for \(10 P_r\):

\(10 P_r = \frac{10!}{(10-r)!}\)

So, we have:

\(\frac{10!}{(10-r)!} = 30240\)

We know that \(10! = 3,628,800\). Substitute this value:

\(\frac{3628800}{(10-r)!} = 30240\)

Now, isolate \((10-r)!\) by rearranging the equation:

\((10-r)! = \frac{3628800}{30240}\)

Performing the division:

\((10-r)! = 120\)

We need to find which factorial equals 120. Let's list the first few factorials:

• \(1! = 1\)
• \(2! = 2 \times 1 = 2\)
• \(3! = 3 \times 2 \times 1 = 6\)
• \(4! = 4 \times 3 \times 2 \times 1 = 24\)
• \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)

We see that \(5! = 120\).

So, we have \((10-r)! = 5!\). This implies:

\(10-r = 5\)

Solving for \(r\):

\(r = 10 - 5\)

\(r = 5\)

Thus, the value of \(r\) is 5.

Final Answer Summary

We solved the permutation equation \(9P_5 + 5 \cdot 9P_4 = 10 P_r\) step-by-step:

• Calculated \(9P_5 = 15120\).
• Calculated \(9P_4 = 3024\).
• Substituted into the equation: \(15120 + 5 \times 3024 = 30240\).
• Set the result equal to \(10 P_r\), so \(10 P_r = 30240\).
• Used the permutation formula \(\frac{10!}{(10-r)!} = 30240\).
• Solved for \((10-r)!\), finding \((10-r)! = 120\).
• Recognized that \(120 = 5!\).
• Set \(10-r = 5\) and solved for \(r\).

The value of \(r\) that satisfies the equation is 5.