Integer Answer Type Questions: Permutations and Combinations | JEE Advanced | 35 Years Chapter wise Previous Year Solved Papers for JEE PDF Download

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 ⁴C₃ ways.
Then one vector from each group can be selected in 2 × 2 × 2 ways.
∴ Total ways = ⁴C₃ × 2 × 2× 2 = 32 = 2⁵  
∴ p = 5

Q.2. Let n₁ < n₂ < n₃ < n₄ < n₅ be positive integers such that n₁ + n₂ + n₃ + n₄ + n₅ = 20. Then the number of such distinct arrangements (n₁, n₂, n₃, n₄, n₅) is   (JEE Adv. 2014)

Ans. (7) 

Sol.  ∵ n₁, n₂, n₃, n₄ and n₅ are positive integers such that n₁ < n₂ < n₃ < n₄ < n₅ 
Then for n₁ + n₂ + n₃ + n₄ + n₅ = 20 If n₁, n₂, n₃, n₄ 
take minimum values 1, 2, 3, 4 respectively then n₅ will be maximum 10.
∴ Corresponding to n₅ = 10,
there is only one solution n₁ = 1, n₂ = 2, n₃ = 3, n₄ = 4.
Corresponding to n₅ = 9, we can have, only solution n₁ = 1, n₂ = 2, n₃ = 3, n₄ = 5 i.e., one solution
Corresponding to n₅ = 8, we can have, only solution n₁ = 1, n₂ = 2, n₃ = 3, n₄ = 6 or n₁ = 1, n₂ = 2, n₃ = 4, n₄ = 4 i.e., 2 solution
For n₅ = 7, we can have n₁ = 1, n₂ = 1, n₃ = 4, n₄ = 6 or n₁ = 1, n₂ = 3, n₃ = 4, n₄ = 5 i.e. 2 solutions
For n₅ = 6, we can have n₁ = 2, n₂ = 3, n₃ = 4, n₄ = 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 = ⁿC₂ – n

∴ ⁿC₂ – n = n ⇒       – 2n = 0

⇒ n² - 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 ⁵C₄ ways. A group of 4 and single girl can be arranged at 2 places out of 6 in ⁶P₂ ways. Also 4 girls can arrange themselves in 4! ways.

∴ m = 5! × ⁶P₂ × ⁵C₄ × 4!