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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:

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

  2. anonymous
    • one year ago
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    This how I understand the question. (2n+1) choose 1 + (2n+1) choose 2 +... +(2n+1) choose n = 63

  3. anonymous
    • one year ago
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    solve for n?

  4. mathmath333
    • one year ago
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    but how u get the equation i didnt understand

  5. mathmath333
    • one year ago
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    yes 3 is the answer

  6. anonymous
    • one year ago
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    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

  7. anonymous
    • one year ago
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    then set equal to 63 . And you can use the identities to simplify this sum

  8. anonymous
    • one year ago
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    once you have solved for n, n is also your maximum number of scholarship candidates

  9. mathmath333
    • one year ago
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    is it something like \(2^{x}-1=63\)

  10. anonymous
    • one year ago
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    if the sum was from i=1 to 2n+1, then yes , it would be 2^(2n+1) -1 its more complicated

  11. anonymous
    • one year ago
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    wolfram gives me a messy expression in terms of Gamma

  12. anonymous
    • one year ago
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    you're supposed to do this by hand? correct

  13. mathmath333
    • one year ago
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    yes

  14. anonymous
    • one year ago
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    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...

  15. mathmath333
    • one year ago
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    the whole thing = 2^(2n+1) -1 right ?

  16. anonymous
    • one year ago
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    yes thats the sum from i=1 to 2n+1

  17. mathmath333
    • one year ago
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    so 2n+1=6,n=2.5 ?

  18. anonymous
    • one year ago
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    its like the pascals triangle, on the way up matches on the way down, relative to the center

  19. anonymous
    • one year ago
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    (2^(2n+1) -1 ) / 2 = 63 , seems like a good approximation

  20. anonymous
    • one year ago
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    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)

  21. anonymous
    • one year ago
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    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

  22. anonymous
    • one year ago
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    (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)

  23. anonymous
    • one year ago
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    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

  24. anonymous
    • one year ago
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    this works because 2n+1 is odd, which produces an even number of terms in the pascal triangle row

  25. anonymous
    • one year ago
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    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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