Permutations and Combinations · Question 17

Let ABC be a triangle. Consider four points p1, p2, p3, p4 on the side AB, five points p5, p6, p7, p8, p9 on the side BC, and four points p10, p11, p12, p13 on the side AC. None of these points is a vertex of the triangle ABC. Then the total number of pentagons, that can be formed by taking all the vertices from the points p1, p2, , p13, is

Let S = \(m, n) : m, n \1, 2, 3, , 50\\. If the number of elements (m, n) in S such that 6m + 9n is a multiple of 5 is p and the number of elements (m, n) in S such that m + n is a square of a prime number is q, then p + q is equal to .

The largest n N, for which 7n divides 101!, is :

1. 18
2. 15
3. 19
4. 16

There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is

1. 230
2. 210
3. 200
4. 220

From a group of 7 batsmen and 6 bowlers, 10 players are to be chosen for a team, which should include at least 4 batsmen and at least 4 bowlers. One batsman and one bowler who are captain and vice-captain respectively of the team should be included. Then the total number of ways such a selection can be made, is

1. 145
2. 165
3. 155
4. 135

For n 2, let Sn denote the set of all subsets of \1, 2, , n\ with no two consecutive numbers. For example \1, 3, 5\ S6, but \1, 2, 4\ S6. Then n(S5) is equal to

Let m and n, (m<n), be two 2-digit numbers. Then the total number of pairs (m, n), such that (m, n)=6, is .

All five letter words are made using all the letters A, B, C, D, E and arranged as in an English dictionary with serial numbers. Let the word at serial number n be denoted by Wn. Let the probability P(Wn) of choosing the word Wn satisfy P(Wn) = 2 P(Wn-1), n > 1. If P(CDBEA) = 2 2 -1, , N, then + is equal to

If the number of seven-digit numbers, such that the sum of their digits is even, is m n 10n; m, n \1, 2, 3, , 9\, then m + n is equal to

Line L1 of slope 2 and line L2 of slope 12 intersect at the origin O. In the first quadrant, P1, P2, , P12 are 12 points on line L1 and Q1, Q2, , Q9 are 9 points on line L2. Then the total number of triangles, that can be formed having vertices at three of the 22 points O, P1, P2, , P12, Q1, Q2, , Q9, is:

1. 1026
2. 1188
3. 1134
4. 1080