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No of ordered triplets of natural number (x,y,z) for which xyz is less than and equal to 11

No of ordered triplets of natural number (x,y,z) for which xyz is less than and equal to 11

Grade:12

1 Answers

Satdhruti Paul
19 Points
6 years ago
You do this in 11 steps.
xyz = 11, this can then be done using the Permutation Combination technique of factoring
xyz = 10, a number into n factors. If a number is of the form N = 2a3b5c.... so on,i.e
xyz = 9,  prime factorise it, then the number of possibilities of xyz = N is given by
.            the answer isa+2c2 * b+2c2 * c+2C2. After finding all of the 11 answers, add them to get your final answer. Hope this helped!
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and so on till
xyz = 2,
xyz = 1. → Here there’s only one solution x = y = z = 1

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