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The value of ${20\choose7}-{20\choose8}+......-{20\choose20}$ is equal to

(a) ${19\choose 13}$

(b) ${19\choose 14}$

(c) ${20\choose 13}$

(d) None of these

My solution:

$(1+x)^{20}={20\choose 0}+{20\choose 1}x+......+{20\choose 20}x^20$

Put $x=-1$

$0={20\choose 0}-{20\choose 1}+......+{20\choose 20}$

${20\choose7}-{20\choose8}+......-{20\choose20}=-{20\choose 0}+{20\choose 1}+....-{20\choose 6}$

And calculated manualy and obtained answer as (a)

My doubt: Is there any other way to solve this?

mathophile
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  • This is an example of "Alternating Sum and Difference of r Choose k up to n". See here: https://proofwiki.org/wiki/Alternating_Sum_and_Difference_of_r_Choose_k_up_to_n – Moeen Nehzati Jan 05 '25 at 18:07
  • ${20\choose7}-{20\choose8}+......-{20\choose20}$ is equal to $-\sum_{k=0}^{13}(-1)^k\binom{20}k$, so your question is similar to: closed form expression for the sum of the first s items of alternating binomial coefficients. If you believe it’s different, please [edit] the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. – Anne Bauval Jan 05 '25 at 18:10
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    You can telescope the sum. You have $\binom{n}{r}=\binom{n-1}{r}+\binom{n-1}{r-1}$. So $\binom{20}{7}=\binom{19}{7}+\binom{19}{6}$, $\binom{20}{8}=\binom{19}{8}+\binom{19}{7}$, etc., and when you put this into your alternating sum everything cancels except $\binom{19}{6}-\binom{19}{20}$. As $\binom{19}{20}=0$ you are left with $\binom{19}{6}$=$\binom{19}{13}$. (Good answers become bad comments when questions get closed :-(.) – Oscar Lanzi Jan 05 '25 at 19:54

1 Answers1

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Once you have reached the answer $$ \binom{20}{0}-\binom{20}{1}\ldots + \binom{20}{6} $$ if you add and subtract $\binom{19}{6}$ you can use property $$ \binom{n+1}{r}-\binom{n}{r}=\binom{n}{r-1}, $$ starting with $\binom{20}{6}-\binom{19}{6}$, and eliminate all terms one by one to be only left with $$ \binom{19}{6} = \binom{19}{13}\quad \mbox{as}\ \underline{the\ answer},\ \mbox{(a) option.} $$

Felix Marin
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Divyam Kapoor
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