Question

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:

Answer 1

\(\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}}\)

Answer 2

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

Answer 3

solve for n?

Answer 4

but how u get the equation i didnt understand

Answer 5

yes 3 is the answer

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

wolfram gives me a messy expression in terms of Gamma

Answer 12

you're supposed to do this by hand? correct

Answer 13

yes

Answer 14

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...

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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)

Answer 21

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

Answer 22

(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)

Answer 23

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

Answer 24

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

Answer 25

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