Problem of the Day
Alicia purchases three different rings that can each be worn on any of her fingers, excluding her thumbs. If she wants to wear at least one ring on each hand, with no more than one ring per finger, how many different ways can she distribute the rings among her eight fingers?
B. 288See the Solution
If it weren’t for the restriction that she has to wear at least one ring on each hand, this would be a straightforward counting problem. There would be $_8P_3=336$ ways for her to wear the rings. However, we don’t want to count the permutations that have her wearing all three rings on the same hand, so we need to subtract those out. For each hand, there are $_4P_3=24$ ways for her to wear all three rings on the same hand. Counting both hands, there are 48 possible ways for her to wear all three rings on the same hand. So, 48 of those 336 permutations calculated above are not allowed, so there are $336-48=288$ ways for her to distribute the rings among her eight fingers.
Another option is to reason as follows:
When she puts on the rings, she has 8 choices for where to put the first ring. If she puts the second ring on the opposite hand of the first ring, she has 4 choices. Now that there is a ring on each hand, she can put the third ring on any of her remaining 6 fingers, so there are $8\cdot 4 \cdot 6=192$ ways for her to distribute the rings in this fashion.
She could put the first ring on any of her 8 fingers, and then put the second ring on the same hand as the first ring, giving her 3 choices. Now, she’s forced to put the third ring on the other hand, giving her 4 choices. There are $8 \cdot 3 \cdot 4=96$ ways for her to distribute the rings in this fashion.
This accounts for all of her options. Either she puts the second ring on the same hand or the opposite hand as the first one. Her choices for the third ring depend directly on what she chooses for the second ring. Adding these two totals together gives us $192+96=288$ ways for her to distribute the rings among her eight fingers.
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