Fun math question...(Light Bulbs)?

Question

There are 10 light bulbs in a box.
You need 4 of them.
How many different combinations of 4 light bulbs are possible?

Note: Knowing how you got your answer would really help..

It's not a trick question.
And the 10*9*8*7 does not work because that assumes that bulb group 10,9,8 and 7 is different than group 7,8,9 and10. But the two groups are the same...

The correct answer is 210, but I did not know how to get it.
NamYzarc explained it very well, and as soon as I can give him/her the "Best Answer" I will...

Answer 1

I assume the 10 light bulbs are all different from each other in some way?

Does the order matter?

If the order DOES matter, the answer is 10*9*8*7, or 5040.

You think of it this way: for your first selection, you pick any of the 10 lightbulbs and put it in slot #1. Then you can pick any of the remaining 9 for slot #2, so that makes 10*9 or 90 different choices. Then 8 for slot #3, 7 for slot #4. Multiply them all together and that's how you get the permutations.

Now, if the order doesn't matter (XYZA is the same as selecting AYZX), you need to divide 5040 by the number of ways you can order 4 items, which ends up being 4*3*2*1, or 24. So 5040/24 is 210. So if you're just picking 4 light bulbs and throwing them into a bag, you can have 210 different combinations in that bag when all's said and done.

Answer 2

You're on the right track. You can have 10 * 9 * 8 * 7 gives you the number of permutations possible. Now, divide that by the number of combinations that can be made from those four objects (the number of ways those objects can be arranged, which is 4!, or 4 * 3 * 2 * 1)
So, the number of combinations that are possible in your question are:
(10 * 9 * 8 * 7) / (4 * 3 * 2 * 1)

This can also be expressed as 10! / (6! * 4 !)

= 210

Answer 3

Hi,
The answer is: 210. You said “combinations,” not arrangements.
First I’ll use the formula for combinations; then I’ll tell you how to do check it with a graphing calculator.
nCr = n!/(r!(n-r))! The number of n objects (10 in your problem), taken r at a time, (4 in your problem.)
So we have:
nCr = 10!/(4!(10-4))!
=(10*9*8*7*!)/(4!*6!)
=(10*9*8*7)/4!
=(10*9*8*7)(4*3*2)
=210 (By all means use your calculator for the arithmetic.)

Now, let’s check it using a calculator. I’ll use a TI-83 Plus, but you can also use most scientific calculators.
a) Enter 10, press MATH, move the cursor to PRB, and press 3. 10nCr will be on the home screen.
b) Enter 4 and press ENTER. The answer 210 will be displayed.

FE

Answer 4

There are 5,040 different combinations.

For the first bulb, you could choose any one of ten. For the second bulb, you could choose any one of nine. Third, you could choose any one of eight. Last, you could choose any one of seven. Multiply 10*9*8*7 -- it should equal 5,040.

I'm thinking this may be some sort of trick question, though, since all ten of the bulbs are presumably the same, and therefore it is impossible to have four DIFFERENT bulbs, unless you mean "different" in the sense that they may be the same STYLE of bulb, but they are not necessarily "unique" bulbs.

Although, maybe they ARE all unique.

Hmmm...

I think I'll stick with my original answer of 5,040.

Answer 5

100

Answer 6

40

Answer 7

answer is 210!

n=10
k=4

c= n!/k!(n-k)!

substituting
c= 10!/4!6!

c=1*2*3*4*5*6*7*8*9*10/1*2*3*4(1*2*3*4*5*6*)

cancelling all like terms
we get

C=7*8*9*10/1*2*3*4

= 5040/24

=210

Answer 8

200

Answer 9

it depends on how many bulbs are different in 10 light bulbs...!?! am i correct? Haha! ;-)

Answer 10

Hundreds, I expect, but I don't know how to work it out! After all we all have pin numbers of four digits and there are thousands of us.