Permutations Practice Test 1

Question : 29 of 40

Marks: +1, -0

There are m points on a straight line AB and n points on another line AC, none of them being the point A. Triangles are formed from these points as vertices when (i) A is excluded (ii) A is included. Then the ratio of the number of triangles in the two cases is:

${m+n-2}/{m+n}$
${m+n-2}/{2}$
${m+n-2}/{m+n+2}$
None of these

Validate

Solution:

Case I: When A is excluded.
Number of triangles = selection of (2 points from AB) (one point from AC) + selection of (one point from AB) (two points from AC)

= ^mC_2 ^nC_1 + ^mC_1 ^nC_2

=1/2(m+n-2)

..... (1)
Case II: When A is included.
The triangles with one vertex at A, one point from AB and one point from AC = mn are to be added in first case.
∴ Number of triangles

=mn+1/2mn(m+n-2)

=1/2mn(m+2)

..... (2)
∴ Required ratio

= (m + n - 2)/(m + n)

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