If 20 C r -20 C r -4- then r C 3 is equal toA- 54B- 56C- 58D- None of these

The correct option is B 56 r + r + 4 = 20 [∵ ^nC_x = ^nC_y ⇒ n = x + y or x = y] ⇒ 2r + 4 = 20 ⇒ 2r = 16 ⇒ r = 8 Now, ^rC_3 = ^8C_3 ^8C_3 = 8!/3! 5! = 8 × 7 × ...

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Standard XII

Mathematics

Combination

If 20 C r =20...

Question

If $^{20} C_{r} =^{20} C_{r + 4},$ then $^{r} C_{3}$ is equal to

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None of these

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Solution

The correct option is B
56

$r + r + 4 = 20$
$[∵^{n} C_{x} =^{n} C_{y} \Rightarrow n = x + y o r x = y]$
$\Rightarrow 2 r + 4 = 20$
$\Rightarrow 2 r = 16$
$\Rightarrow r = 8$
Now, $^{r} C_{3} =^{8} C_{3}$
$^{8} C_{3} = \frac{8!}{3! 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

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If 20Cr=20Cr+4, then rC3 is equal to

The correct option is B 56 r+r+4=20 [∵ nCx=nCy⇒n=x+y or x=y] ⇒2r+4=20 ⇒2r=16 ⇒r=8 Now, rC3=8C3 8C3=8!3!5!=8×7×63×2×1=56

If $^{20} C_{r} =^{20} C_{r + 4},$ then $^{r} C_{3}$ is equal to

The correct option is B
56

$r + r + 4 = 20$
$[∵^{n} C_{x} =^{n} C_{y} \Rightarrow n = x + y o r x = y]$
$\Rightarrow 2 r + 4 = 20$
$\Rightarrow 2 r = 16$
$\Rightarrow r = 8$
Now, $^{r} C_{3} =^{8} C_{3}$
$^{8} C_{3} = \frac{8!}{3! 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$