Question

counting question

Answer 1

\(\large \color{black}{\begin{align}
& \normalsize \text{How many different sums can be formed by the following }\ ?\hspace{.33em}\\~\\
& \normalsize 5\ \text{dollar},\ 1\ \text{dollar},\ 50\ \text{cents},\ 25\ \text{cents},\ 10\ \text{cents},\ 3\ \text{cents},\ 2\ \text{cents},\ 1\ \text{cent.} \hspace{.33em}\\~\\
\end{align}}\)

Answer 2

I guess if you take 2 at a time its 8C2 right?

Answer 3

the 3 at a time it will be 8C3

Answer 4

yes u can also take 3 ,4, 5 upto 8 at a time

Answer 5

yes - but could there be any duplicates in all these?

Answer 6

just need to deduct the double count like example 2 cent and 1 cent taken both are equal to 3 cent taken once

Answer 7

yes exactly

Answer 8

^duplication

Answer 9

and of course there is 8C1 singles

Answer 10

yes

Answer 11

\(\large \color{black}{\begin{align}
& \dbinom{8}{1}+\dbinom{8}{2}+\cdots+\dbinom{8}{8}=2^{8}-1 \hspace{.33em}\\~\\
\end{align}}\)

Answer 12

i think the only duplicates are the ones you mentioned

Answer 13

yes but consider 5 dollar +1cent +2 cent =5 dollar+3cent

Answer 14

this also duplication

Answer 15

right

Answer 16

and 1 dollar + 3 , 1 dollar + 1 + 2

Answer 17

yes all are considered

Answer 18

- also same for 50 , 25, 10 cents

Answer 19

yep

Answer 20

only combinations we are concerned with here right? not permutations

Answer 21

- yes - bacuase we are dealing with sums only

Answer 22

yes only addition (result) is counted

Answer 23

that identity, = 2^8 - 1 I haven't seen that before.

Answer 24

google sum of combinations u will get it

Answer 25

https://www.physicsforums.com/threads/sum-of-combinations.83309/

Answer 26

right
so how do we go about counting the number of duplicates?

Answer 27

yea i m confused about that the main thing

Answer 28

@phi