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I am stuck on the following problem that says ;

If $n \gt 7\,\,$, then prove that $C^{n-1}_3+ C^{n-1}_4 \gt C^{n}_3, \,\, \text{where}\,\, n\in \mathbb{N} $ and $\,\,C^n_r=\frac{n!}{r!(n-r)!}$

My try: For $n \gt 7\,\,$, we check if the statement $P(n): C^{n-1}_3+ C^{n-1}_4 - C^{n}_3 \gt 0\,\,$ is true.

For $n=8$, we see $\,\,P(8): C^7_3+C^7_4-C^8_3 \\ \qquad= \dots\\ \qquad=\frac{8!}{3!.4!}\times \frac{1}{20} \gt 0$.

Similarly, for $n=9,10,\dots$ we can check the veracity of the statement.

Now suppose for $n=m$, the statement is true. That means, $P(m): C^{m-1}_3+ C^{m-1}_4 - C^{m}_3 \gt 0\,\,$. Now , I want to prove the statement is true for $n=m+1$.

Now, $P(m+1): C^{m}_3+ C^{m}_4 - C^{m+1}_3 \\ \qquad \qquad= C^{m+1}_4-C^{m+1}_3$ .

Now, I am stuck. Please show me the right direction.

Air Mike
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learner
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1 Answers1

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By Pascal's rule, $C^{n-1}_3+ C^{n-1}_4 = C^{n}_4\gt C^{n}_3$, since the binomial coefficients in the first half of a row in Pascal's triangle are increasing and $n>7$.

Explicitly, $$ \frac{C^{n}_4}{C^{n}_3} = \frac{n-3}{4} > \frac{7-3}{4} = 1 $$

lhf
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