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Subject: My son's homework One of the problems on my son's, 10 years old, math homework. You have four crayons (red, blue, yellow, green). If you line them up, how many different combinations can you get? Isn't their a way to figure this out without writing all the combinations down? I thought I remembered doing something similar when I took geometry and that there was an equation that could be used to make it easier. Let me know. Neyra Espinoza Hi Neyra, although there are not too many combinations in this case, you definitely want to find quicker methods for determining the number of such arrangements. The easiest way to think about this is to work your way through the problem step by step. In order to make an arrangement you have to start by picking the first crayon. At this point you have four different choices. Suppose you now have made your choice and have put one crayon on the table. For the next crayon in line you once again have to pick one, but now there are only three crayons (hence three choices) left. Combined with the previous four choices you have (so far) a total of 4*3=12 different arrangements. Now you pick the third crayon, but there are only 2 choices left. In total you now have 4*3*2=24 possible arrangements. Finally, the last choice of crayon is not much of a choice, since there's only one crayon left to pick! Hence you have only one choice, and a 4*3*2*1=24 possible arrangements in total. In general, to arrange n different items you will have n! (read "n factorial") = n * (n-1) * (n-2) *... * 3 * 2 * 1 choices.
I hope this helps.
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