Case I When x=1, then y+z=11
There are 4 triplets in this case.
i.e. (1,2,9),(1,3,8),(1,4,7),(1,5,6)
Case II When x=2, then y+z=10
There are 2 triplets in this case i.e. (2,3,7) and (2,4,6).
Case III When x=3, then y+z=9
There is only one triplet in this case i.e. (3,4,5).
∴ Total number of triplets =4+2+1=7
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