Permutation and combination questions

Permutation and Combination questions are an integral part of any aptitude test. The test includes these questions to assess a candidate's ability to recognize patterns and determine links between objects. To solve these questions, one must understand the different types of permutations - the number of combinations possible when arranging a set of objects in a specific order. To get comfortable with numerical aptitude questions, here are some of the most often-asked questions and their answers.

  1. The number of ways of selecting 15 teams from 15 men and 15 women, such that each team consists of a man and a woman is

Answer: Option B

Number of ways selecting 1st team from 15 men and 15 women = \(^{15} C_1 \times ^{15} C_1= 15^2\)
2nd team = \(^{14} C_1 \times ^{14} C_1 =14^2\) and so on
so, total number of ways = \(1^2+2^2+......+15^2= \frac{15 \times 16 \times 31}{6}=1240\)

  1. Let \(A\) and \(B\) be two sets containing four and two elements respectively. Then the number of subsets of the sets \( A \times B\), each having at least three elements is

Answer: Option C

we have \(n(A)=4\) and \(n(B)=2\)
Thus the number of elements in \(A \times B \) is \(8\)
Number of subsets having at least 3 elements
\(=^8C_3 + ^8C_4+^8C_5+^8C_6 +^8 C_7 +^8C_8\)
= \((^8C_0+^8C_1+^8C_2+.....+^8C_8) - (^8C_0+^8C_1+^8C_2)\)
\(= 2^8-(1+8+28)=256-37=219\)

  1. In how many different ways can the letters of the word OPERATE be arranged?

Answer: Option A

The given words contains 8 letters out of which U is taken 2 times and all other letters are different.

∴ Required number of ways =8!2!=8×7×6×5×4×3×2×12


  1. The number of points having both the coordinates as integers, that lie in the interior of the triangle with vertices \((0,0), (0,41), (41,0)\) is

Answer: Option B

We count the number of points on line \(x=n, 1 \leq n < 40, \) that lie in the interior of the triangle .
At line \(x=40\) we have 1 point
\(x=39\) we have 2 points


\(x=n\), we have \((40-n)\) points
The total number of points = \(1+2+3+......+39= \frac{1}{2} \times 39 \times 40 = 39 \times 20=780\)

  1. The sum \( \sum_{r=1} ^{10} (r^2+1) \times (r!)\) is equal to

Answer: Option B

we have, \(T_r= (r^2+1+r-r)(r!)=(r^2+r)(r!)-(r-1)(r!)\)
\(\implies T_r = r (r+1)!-(r-1)r!\)
\(T_1 = 1(2!) -0\)
\(T_2 = 2(3!)- 1(2!)\)
\(T_3= 3(4!) - 2(3!)\)
\(T_{10} = 10(11!) - 9(10!)\)
\( \sum_{r=1} ^{10} (r^2+1) \times (r!) =10 (11!)\)