<p>Hi,</p>
<p>I'm having trouble with a Cambridge bk pg. 489 problem #21:</p>
<p>Three friends are playing a game in which each person simultaneously displays one of three hand signs: a clenched fist, an open palm, or two extended fingers. How many unique combinations of the signs are possible?</p>
<p>A. 3
B. 9
C. 10
D. 12
E. 27</p>
<p>The answer key said that it's C. 10.</p>
<p>Why?</p>
<p>Thanks!</p>
<p>This is why you don’t use non-CB materials. The question is incredibly vague: I can think of more than 10 unique combinations.</p>
<p>Fist Palm fingers 1 unique combo</p>
<p>Fist Fist Palm, Fist Palm Palm, Fist Fist fingers, Fist fingers fingers,
Palm Palm fingers, Palm fingers fingers - another 6 unique combos</p>
<p>Palm Palm Palm, Fingers fingers fingers, fist fist fist - 3 unique combos</p>
<p>(remember, in combos, arrangement of items doesnt matter, only the items themselves matter)
So total 10 combos</p>
<p>Mathematically
A combo would be:
Fist^a . Palm^b . Fingers ^c
Where a + b + c = 3 (Since total 3 people)
Now, if you know how to find number of non-negative solutions of this equation, the answer is clear.
For x1 + x2 + x3 …+ xr = n
the number of non-negative solutions is= n+r-1 C r-1
in our question, n is 3, r is 3… so the answer is 3 + 3 - 1 C 3-1
= 5 C 2 = 5!/2!3! = 20/2 = 10 solutions</p>