1 Crore+ students have signed up on EduRev. Have you? 
Q.1. Consider the set of eight vectors Three non coplanar vectors can be chosen from V in 2p ways. Then p is (JEE Adv. 2013)
Ans. (5)
Sol. Given 8 vectors are (1, 1, 1), (–1, –1, –1); (–1, 1, 1), (1, –1, –1); (1, –1, 1),
(–1, 1, –1); (1, 1, –1), (–1, –1, 1)
These are 4 diagonals of a cube and their opposites.
For 3 non coplanar vectors first we select 3 groups of diagonals and its opposite in ^{4}C_{3 }ways.
Then one vector from each group can be selected in 2 × 2 × 2 ways.
∴ Total ways = ^{4}C_{3} × 2 × 2× 2 = 32 = 2^{5 }
∴ p = 5
Q.2. Let n_{1} < n_{2} < n_{3} < n_{4} < n_{5} be positive integers such that n_{1} + n_{2 }+ n_{3} + n_{4 }+ n_{5} = 20. Then the number of such distinct arrangements (n_{1}, n_{2}, n_{3}, n_{4}, n_{5}) is (JEE Adv. 2014)
Ans. (7)
Sol. ∵ n_{1}, n_{2}, n_{3}, n_{4} and n_{5} are positive integers such that n_{1} < n_{2} < n_{3} < n_{4} < n_{5 }
Then for n_{1} + n_{2} + n_{3 }+ n_{4} + n_{5} = 20 If n_{1}, n_{2}, n_{3}, n_{4 }
take minimum values 1, 2, 3, 4 respectively then n_{5} will be maximum 10.
∴ Corresponding to n_{5} = 10,
there is only one solution n_{1} = 1, n_{2} = 2, n_{3} = 3, n_{4} = 4.
Corresponding to n_{5} = 9, we can have, only solution n_{1} = 1, n_{2 }= 2, n_{3} = 3, n_{4} = 5 i.e., one solution
Corresponding to n_{5} = 8, we can have, only solution n_{1} = 1, n_{2} = 2, n_{3} = 3, n_{4} = 6 or n_{1} = 1, n_{2} = 2, n_{3 }= 4, n_{4} = 4 i.e., 2 solution
For n_{5} = 7, we can have n_{1} = 1, n_{2} = 1, n_{3} = 4, n_{4} = 6 or n_{1} = 1, n_{2} = 3, n_{3} = 4, n_{4} = 5 i.e. 2 solutions
For n_{5} = 6, we can have n_{1} = 2, n_{2} = 3, n_{3} = 4, n_{4 }= 5 i.e., one solution
Thus there can be 7 solutions.
Q.3. Let n ≥ 2 be an integer. Take n distinct points on a circle and join each pair of points by a line segment. Colour the line segment joining every pair of adjacent points by blue and the rest by red. If the number of red and blue line segments are equal, then the value of n is (JEE Adv. 2014)
Ans. (5)
Sol. Number of adjacent lines = n Number of non adjacent lines = ^{n}C_{2} – n
∴ ^{n}C_{2} – n = n ⇒ – 2n = 0
⇒ n^{2}  5n=0 ⇒ n = 0 or 5
But n ≥ 2 ⇒ n = 5
Q.4. Let n be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let m be the number of ways in which 5 boys and 5 girls can stand in a queue in such a way that exactly four girls stand consecutively in the queue.
Then the value of is (JEE Adv. 2015)
Ans. (5)
Sol. n = 5! × 6!
For second arrangement, 5 boys can be made to stand in a row in 5! ways, creating 6 alternate space for girls. A group of 4 girls can be selected in ^{5}C_{4} ways. A group of 4 and single girl can be arranged at 2 places out of 6 in ^{6}P_{2} ways. Also 4 girls can arrange themselves in 4! ways.
∴ m = 5! × ^{6}P_{2} × ^{5}C_{4} × 4!
132 docs70 tests
