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Permutations and Combinations

If r2-13r+40=...

Question

If $r^{2} - 13 r + 40 = 0$ , then the value of $^{7} C_{r}$ is

A

56

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B

12

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C

1

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D

21

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Solution

The correct option is D $21$
Given : $r^{2} - 13 r + 40 = 0$
$\Rightarrow (r - 8) (r - 5) = 0$
$\Rightarrow r = 8, 5$
We know, $n \geq r$ for $^{n} C_{r}$
So, only possible value of $r = 5$
$^{7} C_{r} =^{7} C_{5}$
$\Rightarrow^{7} C_{5} = \frac{7!}{2! \cdot 5!} = \frac{7 \cdot 6 \cdot 5!}{2 \cdot 1 \cdot 5!}$
$∴^{7} C_{5} = 21$

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Similar questions

r^{2} - 13 r + 40 = 0

, then the value of

^{7} C_{r}

^{7} C_{r} \geq {3.}^{7} C_{r - 1}

, then minimum value or

r

^{7} C_{r} + 3^{7} C_{r + 1} + 3^{7} C_{r + 2} +^{7} C_{r + 3} >^{10} C_{4}

, then the quadratic equation whose roots are

α, β

α^{r - 1}, β^{r - 1}

Δ_{r} = ∣ ∣ ∣ ∣ ∣ \begin{matrix} r & 2 r - 1 & 3 r - 2 \frac{n}{2} & n - 1 & a \frac{1}{2} n (n - 1) & (n - 1)^{2} & \frac{1}{2} (n - 1) (3 n - 4) \end{matrix} ∣ ∣ ∣ ∣ ∣

, then the value of

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A = {(r, s) | r < s; r, s \in W}

, then the number of element(s) in set

A

^{7} C_{r} +^{7} C_{r - 1} =^{8} C_{s}

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Permutations and Combinations

Standard XIII Mathematics

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