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In library there are 2n+1 books if a student select at least n+1 books in 256 ways how many number of books in a library?

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There are 9 books
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Maths Tutor

9 books
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There are 9 Books.
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Math Professional Tutor with 16 Years experience

Given that C(2n+1, n+1) + C(2n+1, n+2) + C(2n+1, n+3) +........+ C(2n+1, 2n+1) = 256 --------- (1) => C(2n+1, n) + C(2n+1, n-1) + C(2n+1, n-2) +........+ C(2n+1, 0) = 256 --------- (2) By adding (1) & (2), we get 2^(2n+1) = 512=2^9 => 2n+1 = 9 There fore total number...
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Given that C(2n+1, n+1) + C(2n+1, n+2) + C(2n+1, n+3) +........+ C(2n+1, 2n+1) = 256 --------- (1) => C(2n+1, n) + C(2n+1, n-1) + C(2n+1, n-2) +........+ C(2n+1, 0) = 256 --------- (2) By adding (1) & (2), we get 2^(2n+1) = 512=2^9 => 2n+1 = 9 There fore total number of books in the Library = 9. read less
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Math Professional Tutor with 16 Years experience

Given that C(2n+1, n+1) + C(2n+1, n+2) + C(2n+1, n+3) +........+ C(2n+1, 2n+1) = 256 --------- (1) => C(2n+1, n) + C(2n+1, n-1) + C(2n+1, n-2) +........+ C(2n+1, 0) = 256 ------- (2) By adding (1) & (2), we get 2^(2n+1) = 512=2^9 => 2n+1 = 9 There fore total number...
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Given that C(2n+1, n+1) + C(2n+1, n+2) + C(2n+1, n+3) +........+ C(2n+1, 2n+1) = 256 --------- (1) => C(2n+1, n) + C(2n+1, n-1) + C(2n+1, n-2) +........+ C(2n+1, 0) = 256 ------- (2) [Since C(n, r) = C(n, n-r)] By adding (1) & (2), we get 2^(2n+1) = 512=2^9 => 2n+1 = 9 There fore total number of books in the Library = 9. read less
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There fore total number of books in the Library = 9.
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