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

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n≥4

n>5

n>7

none of these

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Correct option is C

We have, n−1C3+π−1C4>nC3= nC4>nC3 ∵nCr−1+nCr=n+1Cr=n!(n−4)!4!>n!(n−3)!3!⇒14>1n−3⇒n>7

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If n−1C3+n−1C4>nC3, then,

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