Vsolo
Why can't the answer be 4 ordered pairs? (a,b,c) : (4,1,1) and (a,c,b) : (4,1,1)?
Vsolo I can see why you're confused - this is actually a common misconception about ordered triples! Let me clarify what's happening here.
The Key Misunderstanding An ordered triple \((a,b,c)\) is a
single mathematical object consisting of three numbers in a specific order. When we write \( (4,1,1) \), that's ONE ordered triple, not two. The notation \((a,b,c)\) simply means:
\(a\) = first position
\(b\) = second position
\(c\) = third position
These are just
labels for positions, not variables we can rearrange!
Why Your Approach Doesn't Work When you write \((a,b,c): (4,1,1)\) and \((a,c,b): (4,1,1)\), you're not creating two different ordered triples. You're just relabeling the same triple \((4,1,)\) in a confusing way.
Think of it this way: If I have a phone number (555-1234), I can't count it as two different phone numbers by calling the digits different names!
The Correct Solution For \( a^{bc} = 4\), we need to find ALL distinct ordered triples:
Case 1: If \(a = 4\)
\(4^{bc} = 4\) means \(bc = 1\)
Only solution: \(b = 1, c = 1\)
Triple: \( (4,1,1) \)
Case 2: If \(a = 2\)
\(2^{bc} = 4 = 2^2\) means \(bc = 2\)
Solutions: \(b = 1, c = 2\) OR \(b = 2, c = 1\)
Triples: \( (4,2,1) \) and \( (4,1,2) \)
Total: 3 ordered triples
If you'd like, you can practice similar quant questions
here- look for questions involving counting methods, ordered pairs/triples, and exponent properties to reinforce when order creates distinct solutions versus when it doesn't.
Key Takeaway:Each ordered triple is unique based on
which number appears in which position. The triple \( (4,1,1) \) means \(a=4, b=1, c=1\) specifically, and that's just one solution, regardless of how we label the positions.
Remember: In ordered pairs/triples, order matters - \((4,2,1)\) and \((4,1,2)\) are different because the numbers appear in different positions, but \((4,1,1)\) is always just \((4,1,1)\)!