The single digit that are perfect square are 1, 4 and 9, so the units digit of \(A^2\), \(B^2\) and \(C^2\) are 1, 4 and 9.
Which all numbers have their square ending with 1-- 1 and 9
Which all numbers have their square ending with 4-- 2 and 8
Which all numbers have their square ending with 9-- 3 and 7.

(I) Thus A, B and C can be any of 6 but with restriction.
Without restriction we can fill up A,B and C in 6*5*4 = 120 ways.
Now (1 and 9), (2 and 4) and (3 and 7) cannot be together. Let us take them together.
If 1 and 9 are together, the third can be any of the four, so 4*3!=24. Similarly if 3 and 7 are together or 2 and 8 are together, the ways are 4*3!=24, thus 24+24+24=72.
answer 120-72= 48..

(II) Other way
(1,9), (2,8), (3,7)..
ways (1,2,3); (1,2,7); (9,2,3); (9,2,7) and another 4 with 8 instead of 2, that is (1,8,3); (1,8,7); (9,8,3); (9,8,7)
All EIGHT can be arranged in 3! ways so answer is 8*3!=8*6=48..

D
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chetan,

pls elaborate on
If 1 and 9 are together, the third can be any of the four, so 4*3!=24. Similarly if 3 and 7 are together or 2 and 8 are together, the ways are 4*3!=24, thus 24+24+24=72.
answer 120-72= 48..

if 1 and 9 are together, the third number can be selected in 4 ways..( 1 out of 2,8,3,7), and these 3 numbers i.e. 1,9, and the selected number can be arranged in 3! ways. thus total ways = 4*3!.
Similarly for selection when 2 & 8 and 3 & 7 are taken together.

I am sorry, but I don't get why (1 and 9), (2 and 4) and (3 and 7) cannot be together, could you please help me clarify this doubt?

Hello,

Combination will give you only 8 possible selections of A, B and Cs to form the pair (A, B, C). For example, (1,2,3), (1,2,7), (1,8,3), (1,8,7).....
But these 8 possible selections can be rearranged among A, B and Cs --> (2,1,3), (2,1,7), (8,1,3), (7,1,8)..... This rearrangement will give 3! times the original 8 selections.

Solution

Given:

• A, B, and C are three single-digit positive numbers
• The units digit of \(A^2\), \(B^2\), and \(C^2\) are distinct perfect squares

To find:

• The number of possible values of (A, B, C)

Approach and Working:

• The list of all single digit positive numbers are, {1, 2, 3, 4, 5, 6, 7, 8, 9}
• The units digit of squares of these numbers are, {1, 4, 9, 6, 5, 6, 9, 4, 1}
• Among these the perfect squares are {1, 4, 9}

o The numbers whose perfect squares end with 1 are {1, 9}
o The numbers whose perfect squares end with 4 are {2, 8}
o The numbers whose perfect squares end with 9 are {3, 7}

• Thus, the total number of ways of selecting one number from each pair = 2 * 2 * 2 = 8 ways.
• These three numbers can be arranged in 3! = 6 ways

Therefore, total number of values of (A, B, C) = 8 * 3! = 48

Hence the correct answer is Option D.

Answer: D


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