P_1 and P_2 are two distinct and intersecting | vedclass

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(B) Let the set of points on $P_1$ be $S_1 = \{A, B, C\}$ and the set of points on $P_2$ be $S_2 = \{D, E, F\}$.$A$ tetrahedron is formed by selecting

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$15$

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(B) Let the set of points on $P_1$ be $S_1 = \{A, B, C\}$ and the set of points on $P_2$ be $S_2 = \{D, E, F\}$.$A$ tetrahedron is formed by selecting $4$ non-coplanar points.Since $3$ points on $P_1$ are coplanar and $3$ points on $P_2$ are coplanar,we cannot select $4$ points from the same plane.Thus,we must select points from both planes.The possible combinations are:$1$. Select $3$ points from $P_1$ and $1$ point from $P_2$: ${}^3C_3 	imes {}^3C_1 = 1 	imes 3 = 3$.$2$. Select $2$ points from $P_1$ and $2$ points from $P_2$: ${}^3C_2 	imes {}^3C_2 = 3 	imes 3 = 9$.$3$. Select $1$ point from $P_1$ and $3$ points from $P_2$: ${}^3C_1 	imes {}^3C_3 = 3 	imes 1 = 3$.Total number of tetrahedrons $= 3 + 9 + 3 = 15$.

$P_1$ and $P_2$ are two distinct and intersecting planes. Three non-collinear points lie on $P_1$ and another three non-collinear points lie on $P_2$ (none being on the line of intersection of the planes). Then the maximum number of tetrahedrons formed using these six points is:

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