Question

2 questions: 4 bananas are to be selected from a group of 6 how many ways can this be done. 2nd question is the same thing but instead of 6 its 9

Answer 1

bananas all look the same to me :)

Answer 2

gtfo

Answer 3

Serious. You have to pick from distinguishable objects.
Now let us assume the bananas have numbers paintedon them, or something.

Answer 4

Becomes a simple combinations problem. do you know the formula?

Answer 5

nope

Answer 6

if i knew this formula iwouldnt of asked for hel

Answer 7

formula schmormula
pick one banana, there are 6 ways to do it
pick another, there are now 5 choices
counting principle says there are \(6\times 5\) ways to do this altogether, but since you cannot tell the bananas apart you have counted too many ways, since picking banana 1 and then banana 2 is the same as picking banana 2 and then banana 1
therefore you have overcounted by a factor of two and the answer is \(\frac{6\times 5}{2}=3\times 5=15\)

Answer 8

@satellite73 doesn't like formulas
there is something to that.

Answer 9

so what if you can pick four bananas from a group of 9?

Answer 10

pick one banana, there are 9 ways to do it
pick another, there are now 8 choices
counting principle says there are 9×8 ways to do this altogether, but since you cannot tell the bananas apart you have counted too many ways, since picking banana 1 and then banana 2 is the same as picking banana 2 and then banana

The answer is
\[\frac{n!}{(n-k)! \ k!}\]

where n = 9 and k = 4

Answer 11

i didnt get the right answer wtf

Answer 12

I calculate
>>> 9*8*7*6/4*3*2
4536

Answer 13

\[\dbinom{9}{4}=\frac{9\times 8\times 7\times 6}{4\times 3\times 2}=3\times 7\times 6=126\]