# 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

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

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?

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

=20160

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

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

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