## mathmath333 one year ago counting question

1. mathmath333

\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}}

2. phi

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?

3. mathmath333

22C7

4. mathmath333

22C7+25C10

5. phi

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)

6. mathmath333

thnks

7. phi

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

8. mathmath333

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

9. phi

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

10. mathmath333

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

11. phi

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.

12. mathmath333

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

13. mathmath333

ur feedback is valueable.

14. IrishBoy123

"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

15. Zarkon

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.

16. mathmath333

i didn't understand except the first 10 students are choosen from 25 and that 22 comes from the 3 not going from 25