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Say you have 4 apples, 3 strawberries and 6 carrots...

a) If 4 items are to be selected, how many combinations are there?

b) How many combinations will have no carrots?

c) How many combinations will have at least 1 apple?

d) Create the probability distribution table for strawberries.

Now this isn't a simple basic combinatorics formula type question from what I've gathered. I actually did it manually...

a) there are 14 combinations: {a,a,a,a} {a,a,a,s} {a,a,a,c} {a,a,s,s} {a,a,s,c} {a,a,c,c} {a,s,s,s} {a,s,s,c} {a,s,c,c} {a,c,c,c} {s,s,s,c} {s,s,c,c} {s,c,c,c} {c,c,c,c}

b) Again by hand, 13 combinations: {s} {a} {ss} {aa} {sa} {sss} {aaa} {ssa} {saa} {aaaa} {aaaas} {aaaass} {aaaasss}

c) Total number of combinations is 4*3*6=72. To find the ones with at least 1 apple I was thinking of doing total number of combinations and subtracting the combinations with no apples. Combinations with no apples are: {s} {c} {ss} {cc} {cs} {sss} {ccc} {ccs} {css} {cccc} {cccs} {ccss} {csss} {ccccs} {ccccc} {ccccss} {cccccc} {cccccs} {cccsss} {ccccsss} {ccccccc} {ccccccs} {cccccss} {ccccccss} {cccccsss} {ccccccsss}

Therefore 72 – 26 = 46

d) Have not attempted yet, but wondering if there is a better way to do this? My only thoughts are to do it manually but it appears this will take a very long time.

Am I missing something here? It all seems very convoluted. Feedback appreciated.

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# B A simple (?) combinatorics question...

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