SOLUTION: I have a probability problem. If someone can help me figure out the steps and answer, it would be great. Thank you!
problem:
If a mother has three bananas, two pears and two
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Question 425928: I have a probability problem. If someone can help me figure out the steps and answer, it would be great. Thank you!
problem:
If a mother has three bananas, two pears and two oranges, in how many different ways can she give the fruit to her daughter in one week, one piece of fruit per day?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
b = bananas
p = pears
o = oranges
mother has 3b + 2p + 2o
how many different ways can she serve these fruits in one week?
i believe your answer will be 7! / (3! * 2! * 2!) = 210
these are permutations because order is important.
a simple example can be used to show you how this works.
suppose it was only 3 days and you had 2b and 1p
the number of ways would be 3! / (2! * 1!) = 3.
those ways would be:
bba
bab
abb
the first column is the first day, the second column is the second day, the third column is the third day.
a sightly more complex example would be:
2b + 2p over 4 days
answer would be 4! / (2! * 2!) = 6
those ways would be:
bbpp
bpbp
bppb
pbbp
pbpb
ppbb
an even more complex example would be:
2b + 2p + 1o in 5 days.
number of ways would be 5! / (2! * 2! * 1!) = 30
these would be:
bbppo
bbpop
bbopp
bpbpo
bpbop
bppbo
bppob
bpobp
bpopb
pbbpo
pbbop
pbpob
pbpbo
pbopb
pbobp
ppbbo
ppbob
ppobb
obbpp
obpbp
obppb
bobpp
bopbp
boppb
opbbp
opbpb
oppbb
pobbp
pobpb
popbb
it gets harder to show you when the numbers get higher.
trust me on 7! / (3! * 2! * 2!)
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