Imagine that t-shirts are A,B,C,D,E,F.
I think that 20 combination are possible if you don't consider the order, ABC and CAB are only 1 combination.
ABC, ABD, ABE, ABF, ACD, ACE, ACF, ADE, ADF, AEF, BCD, BCE, BCF, BDE, BDF, BEF, CDE, CDF, CEF, DEF.
the formula is: 6! / (3!*(6-3)!)=20
if the boy needed 4 t-shirts: 6! / (4!*(6-4)!)=15
where ! is factorial.....
2007-01-11 13:49:29
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answer #1
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answered by sparviero 6
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Question 1 :
Let 1 be the first t-shirt, 2 be the second and so on.List all the combinations, start with the ones that have t-shirt 1
123
124
125
126
then 1 and 3
134 (132 was above already)
135
136
145
146
156 10 already! Then the remaining with t-shirt 2
234
235
236
245
246
256 6 more!
345
346
356 3 more!
456 1 more!
20 in total!
2007-01-11 21:04:18
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answer #2
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answered by belalondres 1
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1 make a drawing and write all the possibilites out in a list the answer is 20
shirts 1 2 3 4 5 6
123 124 125 126 134 135 136 145 146 156 234 235 236 245 246 256 345 346 356 456
2 1080
2007-01-11 20:57:14
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answer #3
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answered by Anonymous
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I can help you with question #2
The answer is 1080 degrees.
Check out this site for information about how to do this problem:
http://regentsprep.org/Regents/math/poly/LPoly1.htm
As for the first question...you got me. Sorry I couldn't help you. I know that there are 720 combinations for the 6 shirts, but not sure how grouping three together changes this. Sorry :(
2007-01-11 21:00:22
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answer #4
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answered by SteveO 2
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