If n C 3- n C 4- n -1 C 3-then-A- n-7B- None of theseC- n-6 D- n -6

The correct option is C n gt; 6 ^nC_r+^nC_r 1 = ^n+1C_r ^nC_3+^nC_4 ^n+1C_3 ⇒^n+1C_4 ^n+1C_3 ^nC_r/^nC_r 1=n r+1/r ⇒^n+1C_4/^n+1C_3 gt;1⇒n 2/4 gt;1⇒ n ...

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Byju's Answer

Standard XI

Mathematics

Binomial Coefficients

If n C 3+ n C...

Question

If $^{n} C_{3} +^{n} C_{4}$ > $^{n + 1} C_{3},$ then,

n > 6

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n > 7

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n < 6

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None of these

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Solution

The correct option is A
n > 6

$^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$
$^{n} C_{3} +^{n} C_{4} >^{n + 1} C_{3}$
$\Rightarrow^{n + 1} C_{4} >^{n + 1} C_{3}$
$\frac{^{n} C_{r}}{^{n} C_{r - 1}} = \frac{n - r + 1}{r}$
$\Rightarrow \frac{^{n + 1} C_{4}}{^{n + 1} C_{3}}$ > $1 \Rightarrow \frac{n - 2}{4}$ >1 $\Rightarrow$ n > $6.$

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If nC3+nC4>n+1C3,then,

The correct option is A n > 6 nCr+nCr−1=n+1Cr nC3+nC4>n+1C3 ⇒n+1C4>n+1C3 nCrnCr−1=n−r+1r ⇒n+1C4n+1C3>1⇒n−24>1⇒ n >6.

If $^{n} C_{3} +^{n} C_{4}$ > $^{n + 1} C_{3},$ then,