The sides $$AB, BC, CA$$ of a trangle $$ABC$$ have 3,
4 and 5 interior points respectively on them. The total number of triangles that can be constructed by using these points as vertices is
A.
220
B.
204
C.
205
D.
195
Answer :
205
Solution :
We have in all 12 points. Since, 3 points are used to form a triangle, therefore the total
number of triangles including the triangles formed by collinear points on $$AB, BC$$ and $$CA$$ is $$^{12}{C_3} = 220.$$ But this includes the following :
The number of triangles formed by 3 points on $$AB = {\,^3}{C_3} = 1.$$
The number of triangles formed by 4 points on $$BC = {\,^4}{C_3} = 4.$$
The number of triangles formed by 5 points on $$CA = {\,^5}{C_3} = 10.$$
Hence, required number of triangles $$ = 220 - \left( {10 + 4 + 1} \right) = 205.$$
Releted MCQ Question on Algebra >> Permutation and Combination
Releted Question 1
$$^n{C_{r - 1}} = 36,{\,^n}{C_r} = 84$$ and $$^n{C_{r + 1}} = 126,$$ then $$r$$ is:
Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated are
Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4 ; and then the men select the chairs from amongst the remaining. The number of possible arrangements is