A) 120
B) 164
C) 240
D) 82
Correct Answer: A
Solution :
Let \[{{n}_{1}}=1,\]then \[{{n}_{2}}\]can be 2, 3,..., 9 and \[{{n}_{3}}\]can be 3,...., 10 \[\therefore \] No. of ways \[=8+7+6+5+4+3+2+1\]\[=\frac{8\times 9}{2}\] Similarly, when \[{{n}_{1}}=2,\]then \[{{n}_{2}}\] can be 3,..., 9 and \[{{n}_{3}}\] can be 4,...., 10 \[\therefore \]No. of ways\[=7+6+5+4+3+2+1=\frac{7\times 8}{2}\] And so on. \[\therefore \]Total required ways \[=\frac{8\times 9}{2}+\frac{7\times 8}{2}+....+\frac{2\times 3}{2}+\frac{1\times 2}{2}=\frac{240}{2}=120\]You need to login to perform this action.
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