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f(n,k)=n!/(k!*(n-k!))
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31 Oct 2024, 00:11
31 Oct 2024, 00:11
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Question Stats:
99% (01:12) correct
0% (00:00) wrong based on 14 sessions
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\(f(n,k)=\dfrac{n!}{(k!*(n-k!))}\)
Quantity A |
Quantity B |
\(f[16,3]\) |
\(f[16,14]\) |
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f(n,k)=n!/(k!*(n-k!))
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18 Dec 2024, 13:46
18 Dec 2024, 13:46
Expert Reply
We know $\(f(n, k)=\frac{n!}{(k!\times(n-k)!)}\)$; we need to compare $\(f(16,3)\)$ with $\(f(16,14)\)$
Here $\(\mathrm{f}(\mathrm{n}, \mathrm{k}) \equiv{ }^{\mathrm{n \mathrm{C}_{\mathrm{k\)$, so we get column A quantity $\(\mathrm{f}(16,3)={ }^{16} \mathrm{C}_3\)$
As $\({ }^{\mathrm{n \mathrm{C}_{\mathrm{k={ }^{\mathrm{n \mathrm{C}_{\mathrm{n}-\mathrm{k\)$, we get $\(\mathrm{f}(16,14)={ }^{16} \mathrm{C}_{14}={ }^{16} \mathrm{C}_{16-14}={ }^{16} \mathrm{C}_2\)$
Clearly $\(\mathrm{f}(16,3)>\mathrm{f}(16,14)\)$ i.e. column $\(A\)$ quantity is greater than column $\(B\)$ quantity. $\({ }^n C_{r+1}>{ }^n C_r\right\).$, where $\(\left.(r+1)<\frac{n}{2}\right)\)$
Hence the answer is $\((\mathrm{A})\)$.
Here $\(\mathrm{f}(\mathrm{n}, \mathrm{k}) \equiv{ }^{\mathrm{n \mathrm{C}_{\mathrm{k\)$, so we get column A quantity $\(\mathrm{f}(16,3)={ }^{16} \mathrm{C}_3\)$
As $\({ }^{\mathrm{n \mathrm{C}_{\mathrm{k={ }^{\mathrm{n \mathrm{C}_{\mathrm{n}-\mathrm{k\)$, we get $\(\mathrm{f}(16,14)={ }^{16} \mathrm{C}_{14}={ }^{16} \mathrm{C}_{16-14}={ }^{16} \mathrm{C}_2\)$
Clearly $\(\mathrm{f}(16,3)>\mathrm{f}(16,14)\)$ i.e. column $\(A\)$ quantity is greater than column $\(B\)$ quantity. $\({ }^n C_{r+1}>{ }^n C_r\right\).$, where $\(\left.(r+1)<\frac{n}{2}\right)\)$
Hence the answer is $\((\mathrm{A})\)$.
gmatclubot
f(n,k)=n!/(k!*(n-k!)) [#permalink]
18 Dec 2024, 13:46
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