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Find the value of $n$ from the given equation:

$$\binom{n}{n}+2\binom{n}{n-1}+\binom{n}{n-2} = \binom{n+2}{2n-3}$$

I have solved it to get the following and cannot proceed further:

$$\binom{n+2}{2} = \binom{n+2}{2n-3}\\ \frac{(n+2)!}{2!\cdot n!} = \frac{(n+2)!}{(2n-3)!(5-n)!}\\ (2n-3)!(5-n)! = 2!\cdot n!$$

jonsno
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  • There are $0$ ways to choose $n$ elements from a set of $n-1$ elements. (Let alone $n-2$ elements!) – user_194421 Jan 16 '18 at 06:23
  • Wait, isn't $C^{n-1}_{n}=0?$ – Air Conditioner Jan 16 '18 at 06:23
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    For clarification, by $C^r_n$ do you mean the binomial coefficient $\binom{n}{r}=\frac{n!}{r!(n-r)!}$, the number of ways to choose $r$ objects out of $n$ objects total? – JMoravitz Jan 16 '18 at 06:23
  • If the question is to be answered as it is, then $\binom{2n-3}{n+2}=\binom{n}{n}=1$, so $2n-3=n+2\implies n=5$. – user_194421 Jan 16 '18 at 06:25
  • In case this large amount of comments related to confusion about what your $C^r_n$ notation is meant to represent doesn't alarm you, it should. I highly recommend getting away from any notation for binomial coefficients involving $C$ for the very reason that what some people denote as $\binom{n}{r}$ some people denote as $C^r_n$ while others denote as $C^n_r$ and it causes great confusion. The $\binom{n}{r}$ notation is, as far as I know, completely unambiguous as to the meaning. – JMoravitz Jan 16 '18 at 06:28
  • I got to the step: 2.n! = (2n-3)!(5-n)! –  Jan 16 '18 at 06:28
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    @HoàngQuỳnh The comments are asking for the definition of $C^a_b$ which you are using. – jonsno Jan 16 '18 at 06:31
  • Can you show us the steps you've taken? – Air Conditioner Jan 16 '18 at 06:31
  • $\ $ $\ $ $\ $ $C_n^n$ +2$C_n^{n-1}$ + $C_n^{n-2}$ = $C_{n+2}^{2n-3}$

    $\leftrightarrow$ $\ $ $C_{n+1}^n$ + $C_{n+1}^{n-1}$ = $C_{n+2}^{2n-3}$

    $\leftrightarrow$ $\ $ $C_{n+2}^{n}$ = $C_{n+2}^{2n-3}$

    $\leftrightarrow$ $\ $ $\frac{(n+2)!}{2.n!}$ = $\frac{(n+2)!}{(2n-3)!(5-n)!}$

    $\leftrightarrow$ $\ $ 2.n! = (2n-3)!(5-n)!

     –  Jan 16 '18 at 07:08
  • Sorry, because my notation made you confused. I just learned to write like that in school, so I do not know about your notation. –  Jan 16 '18 at 08:23
  • @HoàngQuỳnh $\binom{n}{r} = C_{n}^{r}$ – jonsno Jan 16 '18 at 11:47
  • It mean : choose r in n? –  Jan 16 '18 at 11:54

2 Answers2

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Just $2n-3\leq n+2$ and easy checking.

Michael Rozenberg
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There is another rule (I forgot its name) which says that (for $n, a, b \in \Bbb N$) if:

$$\binom{n}{a} = \binom{n}{b}$$

then either $a = b$ or $n-a = b$.

Apply this rule to your answer $\binom{n+2}{2} = \binom{n+2}{2n-3}$ to get either of the following equations:

  1. $2n-3 = 2$ or

  2. $2n-3 = n$

We reject first one because it gives fractional value for $n$. Only the second one gives valid answer which is $n = 3$.

Of course a smart way (as the smart Michael already posted) is to note that $n+2 \ge 2n-3$ giving $n \le 5$. Checking is very easy then.

jonsno
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