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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 \((4,1,1)\)!
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what is significance of ordered here
Ravixxx
How many ordered triples of positive integers (a,b,c) satisfy \((a^{b})^{c}=4\) ?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
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athakar7 This is actually a crucial distinction that affects our final answer!

Understanding "Ordered" vs "Unordered"

When we say ordered triples, it means the position matters. The triple \((a, b, c)\) represents:
- \(a\) is the first element (the base)
- \(b\) is the second element (the first exponent)
- \(c\) is the third element (the second exponent)

Why This Matters in Our Problem:

The expression \((a^b)^c\) depends on which number goes where. For example:
- \((2, 1, 2)\) means \((2^1)^2 = 2^2 = 4\) ✓
- \((2, 2, 1)\) means \((2^2)^1 = 4^1 = 4\) ✓

Even though both use the same three numbers (\(1, 2, 2\)), they're different ordered triples because the numbers appear in different positions.

If the problem asked for "unordered" sets:
We would count \(\{1, 2, 2\}\) only once, regardless of arrangement. But since it asks for ordered triples, we count \((2, 1, 2)\) and \((2, 2, 1)\) as two distinct solutions.

athakar7
what is significance of ordered here

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