<p>Bob can select one or more of the following 3 toppings for his ice cream: nuts, cream, strawberries. If he selects one or more, how many different combinations of toppings are possible? (assume that the order of the toppings does not matter.)</p>
<p>... Is it 7? if not explain the correct answer to me. Thanks</p>
<p>yea i think ur right</p>
<p>ok good... my friends are indeed incorrect. :p</p>
<p>To break it down (nuts = N, Cream = C, Strawberry = S), since order doesn't matter:</p>
<p>N
C
S
NC
NS
CS
NCS</p>
<p>There are probably formulas for this but its easier to do it in your head.</p>
<p>Yea I showed my friends that and they wouldn't believe me... o well.</p>
<p>3C1+3C2+3C3 = 3+ 3 +1 =7</p>
<p>it's the easiest way to do it,if you have many combinations.</p>
<p>mmm. . .that question is tasty.</p>
<p>2^3 - 1 = 7.
Think of each topping as either being on the icecream cone or off the icecream cone, meaning there are two possibilities for each topping (on or off), and since there are three toppings, there are 2<em>2</em>2 possibilities in total. You have to subtract 1 since we're assuming you order at least one topping.</p>