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Find the value of 𝑟r such that the ratio of 3𝑃𝑟3 P r​ and 4𝑃𝑟−14 P r−1​ will be 1221​ ?

Question

Find the value of r r such that the ratio of 3Pr 3P_r and 4Pr1 4P_{r-1} will be 1221?

3Pr4Pr1=1221 \frac{3P_r}{4P_{r-1}} = 1221

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Solution

1. Break Down the Problem

We need to find the value of r r such that the ratio of 3Pr 3P_r and 4Pr1 4P_{r-1} is equal to 1221.

2. Relevant Concepts

The formula for permutations is given by:

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

Thus, we can express the permutations as:

  • 3Pr=3!(3r)! 3P_r = \frac{3!}{(3-r)!}
  • 4Pr1=4!(4(r1))!=4!(5r)! 4P_{r-1} = \frac{4!}{(4 - (r - 1))!} = \frac{4!}{(5 - r)!}

3. Analysis and Detail

We can set up the equation based on the given ratio:

3Pr4Pr1=1221 \frac{3P_r}{4P_{r-1}} = 1221

Substituting the permutations into the equation:

3!(3r)!4!(5r)!=1221 \frac{\frac{3!}{(3-r)!}}{\frac{4!}{(5-r)!}} = 1221

This simplifies to:

3!(5r)!4!(3r)!=1221 \frac{3! \cdot (5-r)!}{4! \cdot (3-r)!} = 1221

Now, substitute the factorial values:

6(5r)!24(3r)!=1221 \frac{6 \cdot (5-r)!}{24 \cdot (3-r)!} = 1221

Simplifying further gives:

(5r)!4(3r)!=1221 \frac{(5-r)!}{4 \cdot (3-r)!} = 1221

By rewriting (5r)! (5-r)! :

(5r)(4r)(3r)!=12214(3r)! (5-r)(4-r)(3-r)! = 1221 \cdot 4 \cdot (3-r)!

Canceling (3r)! (3-r)! results in:

(5r)(4r)=4884 (5-r)(4-r) = 4884

4. Verify and Summarize

Now let's expand and rewrite the equation:

209r+r2=4884 20 - 9r + r^2 = 4884 r29r+204884=0 r^2 - 9r + 20 - 4884 = 0 r29r4864=0 r^2 - 9r - 4864 = 0

Using the quadratic formula, r=b±b24ac2a r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} , where a=1,b=9,c=4864 a = 1, b = -9, c = -4864 :

  1. Calculate the discriminant: (9)241(4864)=81+19456=19437 (-9)^2 - 4 \cdot 1 \cdot (-4864) = 81 + 19456 = 19437

  2. Calculate r r : r=9±194372 r = \frac{9 \pm \sqrt{19437}}{2} First, calculate 19437 \sqrt{19437} : 19437139.4 \sqrt{19437} \approx 139.4 Therefore, r=9±139.42 r = \frac{9 \pm 139.4}{2}

    This gives two potential values: r1148.4274.2 r_1 \approx \frac{148.4}{2} \approx 74.2 r2130.4265.2(not a valid solution) r_2 \approx \frac{-130.4}{2} \approx -65.2 \quad (\text{not a valid solution})

Final Answer

Thus, the valid integer solution for r r is approximately: r=74 r = 74

This problem has been solved

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