mathmath333
  • mathmath333
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:
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{For a scholarship, at most n candidates out of 2n + 1 can be selected.}\hspace{.33em}\\~\\ & \normalsize \text{ If the number of different ways of selection of at least one candidate}\hspace{.33em}\\~\\ & \normalsize \text{ is 63, the maximum number of candidates that can be selected for } \hspace{.33em}\\~\\ & \normalsize \text{ the scholarship is: } \hspace{.33em}\\~\\ & a.)\ 3 \hspace{.33em}\\~\\ & b.)\ 4 \hspace{.33em}\\~\\ & c.)\ 2 \hspace{.33em}\\~\\ & d.)\ 5 \hspace{.33em}\\~\\ \end{align}}\)
anonymous
  • anonymous
This how I understand the question. (2n+1) choose 1 + (2n+1) choose 2 +... +(2n+1) choose n = 63
anonymous
  • anonymous
solve for n?

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mathmath333
  • mathmath333
but how u get the equation i didnt understand
mathmath333
  • mathmath333
yes 3 is the answer
anonymous
  • anonymous
the 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
  • anonymous
then set equal to 63 . And you can use the identities to simplify this sum
anonymous
  • anonymous
once you have solved for n, n is also your maximum number of scholarship candidates
mathmath333
  • mathmath333
is it something like \(2^{x}-1=63\)
anonymous
  • anonymous
if the sum was from i=1 to 2n+1, then yes , it would be 2^(2n+1) -1 its more complicated
anonymous
  • anonymous
wolfram gives me a messy expression in terms of Gamma
anonymous
  • anonymous
you're supposed to do this by hand? correct
mathmath333
  • mathmath333
yes
anonymous
  • anonymous
there 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
  • mathmath333
the whole thing = 2^(2n+1) -1 right ?
anonymous
  • anonymous
yes thats the sum from i=1 to 2n+1
mathmath333
  • mathmath333
so 2n+1=6,n=2.5 ?
anonymous
  • anonymous
its like the pascals triangle, on the way up matches on the way down, relative to the center
anonymous
  • anonymous
(2^(2n+1) -1 ) / 2 = 63 , seems like a good approximation
anonymous
  • anonymous
you 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
  • anonymous
you 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
  • anonymous
(2n+1) C n = (2n+1) C (n+1) (2n+1) C (n-1) = (2n+1) C (n+2) (2n+1) C (n-2) = (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
  • anonymous
Lets 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
  • anonymous
this works because 2n+1 is odd, which produces an even number of terms in the pascal triangle row
anonymous
  • anonymous
2^(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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