If r 2-13 r -40-0- then the value of 7 C r isA- 1B- 21- 56D- 12

The correct option is B 21Given : r^2 13r+40=0 ⇒ (r 8)(r 5)=0 ⇒ r=8,5 We know, n≥r for ^nC_r nbsp; So, only possible value of r=5 ^7C_r= ^7C_5 ⇒ ^7C_5=7!2!·5 ...

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Standard XII

Mathematics

Permutations and Combinations

If r 2-13 r +...

Question

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

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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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If r2−13r+40=0, then the value of 7Cr is

The correct option is D 21Given : r2−13r+40=0 ⇒ (r−8)(r−5)=0 ⇒ r=8,5 We know, n≥r for nCr So, only possible value of r=5 7Cr= 7C5 ⇒ 7C5=7!2!⋅5!=7⋅6⋅5!2⋅1⋅5! ∴ 7C5=21

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