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- 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:

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- mathmath333

- schrodinger

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- 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

This how I understand the question.
(2n+1) choose 1 + (2n+1) choose 2 +... +(2n+1) choose n = 63

- anonymous

solve for n?

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- mathmath333

but how u get the equation i didnt understand

- mathmath333

yes 3 is the answer

- 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

then set equal to 63 .
And you can use the identities to simplify this sum

- anonymous

once you have solved for n, n is also your maximum number of scholarship candidates

- mathmath333

is it something like \(2^{x}-1=63\)

- anonymous

if the sum was from i=1 to 2n+1, then yes , it would be 2^(2n+1) -1
its more complicated

- anonymous

wolfram gives me a messy expression in terms of Gamma

- anonymous

you're supposed to do this by hand? correct

- mathmath333

yes

- 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

the whole thing = 2^(2n+1) -1 right ?

- anonymous

yes thats the sum from i=1 to 2n+1

- mathmath333

so 2n+1=6,n=2.5 ?

- anonymous

its like the pascals triangle, on the way up matches on the way down, relative to the center

- anonymous

(2^(2n+1) -1 ) / 2 = 63 , seems like a good approximation

- 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

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

(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

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

this works because 2n+1 is odd, which produces an even number of terms in the pascal triangle row

- 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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