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I got this question and I am little bit confused, whether the question is correct or not.

$n\choose0$-$ n\choose1 $+$ n\choose2$-$ n\choose3$+.........+$(-1)^r$$ n\choose r$=$28$

Now we are required to find the values of n. Its obvious that its an incomplete expression of $(1-x)^n$ but if we don't know the $r$ how can we go on to find $n$. The answer was $n=9$ but actually it would be worthless if we don't know the values where we have to stop i.e the value of $r$.

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    $\begingroup$ In this type of question, you are required to find both $r$ and $n$, but you only have to report the value(s) of $n$. $\endgroup$ – Strategy Thinker Jun 18 '16 at 16:10
  • $\begingroup$ @StrategyThinker Is there any direct way in which I can find both the values or do I have to to use the put and check approach on the question by putting the values in the expression $(-1)^r $${ n-1}\choose{r}$ $\endgroup$ – Harsh Sharma Jun 18 '16 at 16:15
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That's because $$ \eqalign{ & \sum\limits_{0\, \le \,k\, \le \,r} {\left( { - 1} \right)^{\,k} \left( \matrix{ n \cr k \cr} \right)} = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,r} \right)} {\left( { - 1} \right)^{\,k} \left( \matrix{ r - k \cr r - k \cr} \right)\left( \matrix{ n \cr k \cr} \right)} = \cr & = \left( { - 1} \right)^{\,r} \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,r} \right)} {\left( \matrix{ - 1 \cr r - k \cr} \right)\left( \matrix{ n \cr k \cr} \right)} = \left( { - 1} \right)^{\,r} \left( \matrix{ n - 1 \cr r \cr} \right) \cr} $$ after which $$ 28 = 7 \cdot 4 = {{8 \cdot 7} \over {1 \cdot 2}}\quad \to \quad \left( {n = 9,\;r = 2} \right)\; \vee \;\left( {n = 9,\;r = 6} \right) $$

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The expression $\sum_{k=0}^{r}{n \choose k}{(-1)}^k$ can be simplified as ${(-1)}^r{n-r \choose r}$. Skipping $r=0,1$, you can see that $r$ is even and for $r \geq 4$ there is no solution, alternatively see where 28 appears in Pascal's triangle.

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  • $\begingroup$ That's actually a type of put and check situation. Is there any way where we can get to the answer through the systematic approach. $\endgroup$ – Harsh Sharma Jun 18 '16 at 16:10
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    $\begingroup$ should be ..Bin(n-1,r), if I do not mistake $\endgroup$ – G Cab Jun 18 '16 at 16:10
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    $\begingroup$ @HarshSharma You could try $r=0$, simplify the expression in terms of $n$ and try solve for $n$. Then try $r=1$, which gives a new expression in terms of $n$. Once you do this for various values of $r$, you will notice the above formula. $\endgroup$ – Strategy Thinker Jun 18 '16 at 16:14

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