Question

counting question

Answer 1

\(\normalsize \color{black}{\begin{align}
& \text {From a class of 25 students, 10 are to be chosen for an excursion party. }\hspace{.33em}\\~\\
& \text{ There are 3 students who decide that either all of them will join or none}\hspace{.33em}\\~\\
& \text{ of them will join. In how many ways can the excursion party be chosen ?}\hspace{.33em}\\~\\
\end{align}}\)

Answer 2

divide the class into 3 and 22
if we choose the 3, then there are 7 more left to choose
if we don't choose the 3 we must choose 10
can you post your try?

Answer 3

22C7

Answer 4

22C7+25C10

Answer 5

that is if we choose the 3
if we don't, then we have another set of choices
I would add them

22C7 + 22C10

(the first includes the "3" group)

Answer 6

thnks

Answer 7

notice the second is not 25C10 (because this would allow us to choose one of the "3" and they are all together)

Answer 8

i particularly don't understand the language of this world problems, except only when some one explains me

Answer 9

this stuff is confusing, though if you do it enough (and that means *a lot*)
it eventually gets a (little) clearer.

Answer 10

as there is less time in exam, i get panicked to see even simple questions

Answer 11

two thoughts:
1) you are smarter than other people so they will do even worse
2) if you are confused by a question (do this when studying), ask what is confusing?
when you get the answer, try to figure out why the answer makes sense, and why it did not before you got the answer.

Answer 12

yes u are right i need to practice a lot at least 100 questions on my own

Answer 13

ur feedback is valueable.

Answer 14

"i particularly don't understand the language of this world problems"

you're not the only one !!

here, assumptions are made about how this plays out, including the following:

1) the decision only to go together is communicated to decision makers [trivial]
2) the decision is accepted by decision makers [important] who then decide to treat *equally* [just as important] options a) and b) where
a) = sending all 3, plus 7 others from the remaining 22 and
b) = choosing 10 from the 22.

if it was about pulling balls out of a hat, so you can only draw 10 from 25, but 3 are glued together [yet count as 3], i think i would it might seem easier. certainly less hand-wavey

Answer 15

you can also do

\[{25\choose 10}-{22\choose8}{3\choose2}-{22\choose9}{3\choose1}\]

See if you can figure out why this is the same.

Answer 16

i didn't understand except the first 10 students are choosen from 25

and that 22 comes from the 3 not going from 25