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Find all $a,b,c \in \Bbb N $ such that

$$\binom{a}{b} \binom{b}{c}=2\binom{a}{c}$$

$(c\leq b \leq a)$

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  • $\begingroup$ Use the definition of $\binom{m}{n}$, and simplify. $\endgroup$ – Arthur Dec 16 '14 at 11:49
  • $\begingroup$ Alternatively, think of what the binomial coefficients count (i.e. how many ways of choosing $b$ items from a set of size $a$) $\endgroup$ – Kris Dec 16 '14 at 11:51
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Let $a-c=m$ and $a-b=n$ in simmons's answer. Then you have, $$\binom{m}{n}=2 \implies \dfrac{m!}{n!(m-n)!}=2\implies \dfrac{(m-n+1)(m-n+2)\cdots m}{n!}=2$$

$m=n+k \implies \dfrac{(k+1)(k+2)\cdots (n+k)}{n!}=2\tag{1}$

$k\geq 1 \implies \dfrac{(k+1)(k+2)\cdots (n+k)}{n!}\geq {n+1}\geq1$

If $n>1$ then there exists no solution for $(1)$. Therefore $n=1$. But if $n=1$ then $k=1$ so that a solution of $(1)$ exists.

Thus we have,

$$a-b=1, a-c=2$$

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Your equation simplifies to $$\binom{a-c}{a-b}=2.$$ Can you deduce the answer from here?

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  • $\begingroup$ How can i deduce ? $\endgroup$ – erfan soheil Dec 16 '14 at 13:28
  • $\begingroup$ @erfansoheil Have you tried using the definition of $\binom{m}{n}$ to deduce this? $\endgroup$ – simmons Dec 16 '14 at 15:19

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