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mathmath333
 one year ago
For a scholarship, at most n candidates out of 2n + 1 can be selected.
If the number of different ways of selection of at least one candidate
is 63, the maximum number of candidates that can be selected for
the scholarship is:
mathmath333
 one year ago
For a scholarship, at most n candidates out of 2n + 1 can be selected. If the number of different ways of selection of at least one candidate is 63, the maximum number of candidates that can be selected for the scholarship is:

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0\(\large \color{black}{\begin{align} & \normalsize \text{For a scholarship, at most n candidates out of 2n + 1 can be selected.}\hspace{.33em}\\~\\ & \normalsize \text{ If the number of different ways of selection of at least one candidate}\hspace{.33em}\\~\\ & \normalsize \text{ is 63, the maximum number of candidates that can be selected for } \hspace{.33em}\\~\\ & \normalsize \text{ the scholarship is: } \hspace{.33em}\\~\\ & a.)\ 3 \hspace{.33em}\\~\\ & b.)\ 4 \hspace{.33em}\\~\\ & c.)\ 2 \hspace{.33em}\\~\\ & d.)\ 5 \hspace{.33em}\\~\\ \end{align}}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This how I understand the question. (2n+1) choose 1 + (2n+1) choose 2 +... +(2n+1) choose n = 63

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0but how u get the equation i didnt understand

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0yes 3 is the answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the number of ways to choose 1 candidate from 2n+1 candidates is (2n+1) choose 1 the number of ways to choose 2 candidates from 2n+1 candidates is (2n+1) choose 2 ... the number of ways to choose n candidates from 2n+1 candidates is (2n+1) choose n and the directions say find the number of ways to choose at least 1, but the maximum is n so add the above

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then set equal to 63 . And you can use the identities to simplify this sum

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0once you have solved for n, n is also your maximum number of scholarship candidates

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0is it something like \(2^{x}1=63\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if the sum was from i=1 to 2n+1, then yes , it would be 2^(2n+1) 1 its more complicated

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0wolfram gives me a messy expression in terms of Gamma

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you're supposed to do this by hand? correct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there is a symmetry (2n+1) choose 1 = (2n+1) choose ( 2n+1 1) (2n+1) choose 2 = (2n+1) choose ( 2n+1 2) etc...

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.0the whole thing = 2^(2n+1) 1 right ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes thats the sum from i=1 to 2n+1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its like the pascals triangle, on the way up matches on the way down, relative to the center

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(2^(2n+1) 1 ) / 2 = 63 , seems like a good approximation

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can use the fact that there is going to be an even number of terms in the pascals triangle, because 2n+1 is odd (and you add one extra value for the zeroth term)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you can get familiarize with binomials by looking at 7 7 C 0 + 7 C 1 + 7 C 2 + 7 C 3 7 C 4 + 7 C 5 + 7 C 6 + 7 C 7 7 C 3 = 7 C 4 7 C2 = 7 C 5 7 C 1 = 7 C 6 7 C 0 = 7 C 7

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0(2n+1) C n = (2n+1) C (n+1) (2n+1) C (n1) = (2n+1) C (n+2) (2n+1) C (n2) = (2n+1) C (n+3) ... (2n+1) C 1 = (2n+1) C (2n) notice that what is missing from both is (2n+1) C 0 = (2n+1) C (2n+1)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Lets sum up the binomial terms from i=0 to i= 2n+1, divide it in half (since we only want half the terms) then get rid of the zeroth term 2^(2n+1)/ 2  (2n+1) C 0 = 63 this will give you n exactly

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this works because 2n+1 is odd, which produces an even number of terms in the pascal triangle row

anonymous
 one year ago
Best ResponseYou've already chosen the best response.02^(2n+1)/ 2  (2n+1) C 0 = 63 2^(2n+1) /2  1 = 63 2^(2n+1) /2 = 64 2^(2n+1) = 128 n = 3
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