If 12Pr=11P6+6⋅11P5 then r is equal to :

Given : 12Pr=11P6+611P5We know that n−1Pr+rrn−1Pr−1=r!nCr∴ 11P6+6.11P5=6!12C6⇒ 12P6=6!12C6∴ 12!6!=6!12!6!6! which are equal ∴ r=6

If 12Pr=11P6+6⋅11P5 then r is equal to :

Given : 12Pr=11P6+611P5We know that n−1Pr+rrn−1Pr−1=r!nCr∴ 11P6+6.11P5=6!12C6⇒ 12P6=6!12C6∴ 12!6!=6!12!6!6! which are equal ∴ r=6

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If $^{12} P_{r} =^{11} P_{6} + 6 \cdot^{11} P_{5}$ then $r$ is equal to :

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Detailed Solution

Given : $^{12} P_{r} =^{11} P_{6} + 6^{11} P_{5}$

We know that

$\begin{array}{ll} ^{n - 1} P_{r} + r r^{n - 1} P_{r - 1} = r!^{n} C_{r} \\ ∴^{11} P_{6} + 6 .^{11} P_{5} = 6!^{12} C_{6} \\ \Rightarrow^{12} P_{6} = 6!^{12} C_{6} \\ ∴ \frac{12!}{6!} = 6! \frac{12!}{6! 6!} which are equal \\ ∴ r = 6 \end{array}$