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प्रश्न
How many triangles can be formed by 15 points, in which 7 of them lie on one line and the remaining 8 on another parallel line?
उत्तर
To form a triangle we need 3 non-collinear points.
Take the 7 points lying on one line be group A and the remaining 8 points lying on another parallel line be group B.
We have the following possibilities
| Group A 7 points |
Group B 8 points |
Combination | |
| (i) | 2 | 1 | 7C2 × 8C1 |
| (ii) | 1 | 2 | 7C1 × 8C2 |
∴ Required number of ways of forming the triangle
= (7C2 × 8C1) + (7C1 × 8C2)
= `(7!)/(2!(7 - 2)!) xx 8 + 7 xx (8!)/(2!(8 - 2)!)`
= `(7!)/(2 xx 5!) xx 8 + 7 xx (8!)/(2! xx 6!)`
= `(7 xx 6 xx 5! xx 8)/(2! xx 5!) + (7 xx 8 xx 7 xx 6!)/(2! xx 6!)`
= `(7 xx 6 xx 8)/(2! xx 5!) + (7 xx 8 xx xx 6!)/(2! xx 6!)`
= `(7 xx 6 xx8)/(2 xx 1) + (7 xx 8 xx 7)/(2 xx 1)`
= 7 × 6 × 4 + 7 × 4 × 7
= 168 + 196
= 364
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