I have 3 pieces of candy to place in 4 lunch boxes. In how many ways can I do this if: (a) The candies are all the same and the
lunch boxes are all different? (b) The candies are all different and the lunch boxes are all the same? (c) Exactly two of the candies are the same (but the third is different) and all of the lunch boxes are different? Please explain your work IN DETAIL. Thanks!
<span>I have 3 pieces of candy to place in 4 lunch boxes. In how many ways can I do this if: (a) The candies are all the same and the lunch boxes are all different? (b) The candies are all different and the lunch boxes are all the same? (c) Exactly two of the candies are the same (but the third is different) and all of the lunch boxes are different? Please explain your work IN DETAIL. Thanks!?.</span>
<span>A).4^3 = 64
</span><span>B).(3+4-1)C3 = 6c3 = 20
</span><span>C).3c3 +3c2 + 3c1 = 7
</span><span>The last Question). 4!/2! ways = 4+24+12 = 40 ways </span> Hoped This Helped, <span>
Alialbaghdadi6 Your Welcome :) </span>
PEMDAS Solve the parentheses first so 9+4=13. Then do the division so 6/6=1. Now the equation is 37- 13+ 1. Then work left to right. 37-13= 24 then 24+1=25.
