Multistep combinational problem (binomial expansion)

[niceboar]

Homework Statement

There are 30 calculators and a teacher selects a group of 25 of them. 4 of them need to be recharged. How many samples of the 25 calculators contain all 4 of the calculators that need to be recharged

a. 30C4
b. 25C4
c. 30C25
d. 26C21
e. none

Homework Equations

nCr = n!/(r!(n!-r!))

The Attempt at a Solution

Well the problem is that I can't make much sense of this.

30C25 would be the total number of combinations of the 25 size group of calculators
25C4 would be how many groups of 4 that could be made from 25 calculators. Since we are looking for 1 group and only one in particular that fits the parameters there must be 1/25C4

but I can't figure out how that's related to the total number of groups 30C25

or could it be
30C25 is the total number of groups

we need to select 4 calculators that needs to be recharged and 21 others

so

25C21 * 25C4 = ways of choosing a group of 25 with our 4 needs to be recharged

I'm really lost here

 

[vela]

Think of it this way. You know 4 of the 25 have to be the ones that need recharging. That leaves 21 spots you have to fill. How many calculators do you have to choose from to fill the 21 spots?

 

[niceboar]

Think of it this way. You know 4 of the 25 have to be the ones that need recharging. That leaves 21 spots you have to fill. How many calculators do you have to choose from to fill the 21 spots?

so 21 blank slots 4 are accounted for already (or the combination doesn't count) in the 25 so I would need to choose 21 calculators from 26 leftover?

26C21? Thank you.

 

[vela]

Exactly.

You could also look at the 5 calculators that aren't chosen. If none of those needs recharging, then the 4 that do must be in the 25. So how many ways can you choose 5 calculators that don't need to be recharged? You'll come up with a different expression than 26C21, but you should be able to show the two answers are in fact equal.

 

[niceboar]

Exactly.

You could also look at the 5 calculators that aren't chosen. If none of those needs recharging, then the 4 that do must be in the 25. So how many ways can you choose 5 calculators that don't need to be recharged? You'll come up with a different expression than 26C21, but you should be able to show the two answers are in fact equal.

26C5 should equal 26C21 although looking at it that way seems more confusing

 

[niceboar]

so part 2 of this problem is

of the 25 how many groups have at least 1 calculator that needs to be recharged

so I need 1 out of the 25 + 2/25 etc

so

25C1 * 30C24 + 25C2 * 30C23 + 25C3 * 30C22 + 25C4 * 30C21

? since I'd choose 1 out of 25 to be broken then choose the remaining 24 to be OK then add the results til 4?

wait this doesn't seem to be right the number would be really large

think this would be total ways?

 

[niceboar]

oh would I do

29C24 + 28C23 + 27C22 + 26C21?

 

[niceboar]

doesn't seem like it'd be right either

looking at the answers

30C25 -> choose any 25 of the 30
25C4 -> all combinations of the 4 in the group of 25

so would it be 30C25 - 25C4? This isn't exactly an answer but there is 30C25 - 26C4 and 30C25 - 26C25

30C25 -> choose any 25
30-4 = 26 so 26C4 doesn't seem to make sense
30C25 -> choose any 25
26C25 -> of the 26 good calculators choose 25

30C25 - 26C25 -> not sure really how to interpret a negative here
choose any 25 of the 30 then subtract the number of combinations of ONLY working calculators = combinations of at least one non working calculator? I am thinking this is the correct answer

 

[vela]

Sorry, I didn't realize you had posted more.

You're now selecting 25 at random and want to know how many of the possible groups have at least 1 calculator that needs to be recharged, right? In this case, it's easier to calculate how many groups there are which contain no calculators that need to be recharged. Then subtract that number from the number of all possible groups.

This is because the complement of "at least 1" is "none".