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If .^(n-1)P3 :^(n+1)P3 = 5 : 12, find n....

If `.^(n-1)P_3 :^(n+1)P_3 = 5 : 12,` find `n`.

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To solve the problem, we need to find the value of \( n \) given the ratio \( \frac{(n-1)P_3}{(n+1)P_3} = \frac{5}{12} \). ### Step-by-Step Solution: 1. **Understand the formula for permutations**: The formula for permutations is given by: \[ nPr = \frac{n!}{(n-r)!} \] For our case, we have \( r = 3 \). 2. **Write the expressions for \( (n-1)P_3 \) and \( (n+1)P_3 \)**: \[ (n-1)P_3 = \frac{(n-1)!}{(n-1-3)!} = \frac{(n-1)!}{(n-4)!} \] \[ (n+1)P_3 = \frac{(n+1)!}{(n+1-3)!} = \frac{(n+1)!}{(n-2)!} \] 3. **Set up the ratio**: \[ \frac{(n-1)P_3}{(n+1)P_3} = \frac{\frac{(n-1)!}{(n-4)!}}{\frac{(n+1)!}{(n-2)!}} = \frac{(n-1)! \cdot (n-2)!}{(n+1)! \cdot (n-4)!} \] 4. **Simplify the ratio**: \[ = \frac{(n-1)! \cdot (n-2)!}{(n+1) \cdot n \cdot (n-1)! \cdot (n-4)!} \] Canceling \( (n-1)! \) from numerator and denominator: \[ = \frac{(n-2)!}{(n+1) \cdot n \cdot (n-4)!} \] 5. **Further simplify**: \[ = \frac{(n-2)(n-3)(n-4)!}{(n+1) \cdot n \cdot (n-4)!} \] Cancel \( (n-4)! \): \[ = \frac{(n-2)(n-3)}{(n+1) \cdot n} \] 6. **Set the equation equal to the given ratio**: \[ \frac{(n-2)(n-3)}{(n+1) \cdot n} = \frac{5}{12} \] 7. **Cross-multiply to eliminate the fraction**: \[ 12(n-2)(n-3) = 5(n+1)n \] 8. **Expand both sides**: - Left side: \[ 12(n^2 - 5n + 6) = 12n^2 - 60n + 72 \] - Right side: \[ 5(n^2 + n) = 5n^2 + 5n \] 9. **Set the equation to zero**: \[ 12n^2 - 60n + 72 = 5n^2 + 5n \] Rearranging gives: \[ 12n^2 - 5n^2 - 60n - 5n + 72 = 0 \] \[ 7n^2 - 65n + 72 = 0 \] 10. **Use the quadratic formula to find \( n \)**: The quadratic formula is given by: \[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 7, b = -65, c = 72 \): \[ n = \frac{65 \pm \sqrt{(-65)^2 - 4 \cdot 7 \cdot 72}}{2 \cdot 7} \] \[ = \frac{65 \pm \sqrt{4225 - 2016}}{14} \] \[ = \frac{65 \pm \sqrt{2209}}{14} \] \[ = \frac{65 \pm 47}{14} \] 11. **Calculate the two possible values for \( n \)**: - First value: \[ n = \frac{112}{14} = 8 \] - Second value: \[ n = \frac{18}{14} \approx 1.28 \] 12. **Determine the valid solution**: Since \( n \) must be a natural number, the only valid solution is: \[ n = 8 \] ### Final Answer: The value of \( n \) is \( 8 \).
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NAGEEN PRAKASHAN ENGLISH-PERMUTATION AND COMBINATION -Exercise F
  1. If .^(16)C(r) = .^(16)C(r+6), then find .^(5)C(r).

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  2. Determine n if (i) ^2n C2:^n C2=12 :1 (ii) ^2n C3:^n C3=11 :1

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  3. Determine n if (i) .^(2n) C2: "^n C2=12 :1 (ii) .^(2n) C3: "^n C...

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  4. Determine n if (i) ^2n C2:^n C2=12 :1 (ii) ^2n C3:^n C3=11 :1

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  5. If .^(15)C(r): .^(15)C(r-1) = 1:5, then find r.

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  6. If .^(n-1)P3 :^(n+1)P3 = 5 : 12, find n.

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  7. If .^(n)P(r) = 720 and .^(n)C(r) = 120, then find r. .

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  8. If .^(n+1)C(r+1.): ^nCr: ^(n-1)C(r-1)=11:6:3 find the values of n and ...

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  9. If ""^(n)C(4),""^(n)C(5) and ""^(n)C(6) are in A.P. then the value of ...

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  10. If alpha=\ \ ^m C2,\ then find the value of \ ^(alpha)C2dot

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  11. In how many ways can a team of 11 players be selected from 14 players?

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  12. In how many ways 2 persons can be selected from 4 persons?

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  14. In how many ways can 11 players be selected from 14 players if (i) a...

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  15. In how many ways can 5 subjects be chosen from 9 subjects if three sub...

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  16. In how many ways can 4 books be chosen from 12 books if (i) there is...

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