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Question

The number of ways in which 57 can be expressed as a product of three factors:

A
7
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B
8
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C
9
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D
none of these
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Solution

The correct option is B 8
57=5a×5b×5c
Such that a+b+c=7 where a, b, c are all whole numbers.
Now we know that a+b+c=7 has 7+31C2=9C2=36 number of solutions.
But 36 incluces those solutions which have any 2 variables equal to each other. Let's check for those. Let us consider a= b
then 2a + c = 7
For the above equation, the number of solutions is 4. But since any 2 variables can be taken therefore total number of such cases = 3C2×4=12
Removing these cases and keeping in mind that we have to find out the number of ways in an unordered form, we get 36126+4=4+4=8
Hence, total number of ways = 8
{0,0,7},{0,4,3},{0,6,1},{0,5,2},{1,4,2},{1,3,3},{1,1,5},{2,2,3}

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