LTI Permutation and combinations Quiz-1

We know if n student sit around a circular table Number of possible Arrangements will be (n -1)!
Then (9-1)! = 8! = 40320

In a password At thousands place there can be 0 but in Numbers we can not keep 0 in thousands place.
At unit’s place possible digits = 10
At ten’s place possible digits = 10
At hundreds place possible digits = 10
At thousands place possible digits =9

Number of possible combinations = 10*10*10*9 = 9000

7 Married couples means we have to arrange 14 peoples
First women can be placed anywhere in a circular Round Table and her husband in 1 way.
Second women can be placed in 12 ways and her husband in 1 way.
Third woman can be placed in 10 ways and her husband in 1 way.
Fourth women can be placed in 8 ways and her husband in 1 way.
Fifth women can be placed in 6 ways and her husband in 1 way.
And so on,
Total number of ways 12*10*8*6*4*2 = 46080 Ways

For a rectangle we need to select 2 parallel lines and the other two parallel lines that are perpendicular to each other.
Hence, ^{3}\textrm{C}_{2}\times ^{6}\textrm{C}_{2} = 45 rectangles

As whenever any two horizontal and any two vertical lines are chosen their intersection will
produce a rectangle.
Number of Rectangles = ^{3+1}\textrm{C}_{2}\times ^{4+1}\textrm{C}_{2} = 360
Number Of Rectangle will be 60

At unit’s place possible digits = 3 which is 0, 2 ,8

When at unit’s place 0 then
Hundred place possible digits = 7
Ten’s place possible digits = 6
Total = 6 x 7 =42

When at unit’s place 2 then
Hundred place possible digits = 6
Ten’s place possible digits = 6
Total = 6 x 6 =36

When at unit’s place 8 then
Hundred place possible digits = 6
Ten’s place possible digits = 6
Total = 6x 6 =36

Possible Even Numbers = 42+36+36= 114

The 5 letter word can be rearranged in 5!=120 Ways without any of the letters repeating.
The first 24 of these words will start with A.
Then the 25th word will start EA _ _ _.
The remaining 3 letters can be rearranged in 3!=6 Ways. i.e. 6 words exist that start with EA.
The next word starts with EG and then A, i.e., EGA _ _.
The first of the words will be EGART. The next word will be EGART.
Therefore, the rank of EGART will be 24+6+2= 32

Number of arrangements possible
[/latex]=\frac{1}{2} \times (9-1)!=\frac{1}{2} \times 8!=\frac{1}{2} \times40320=20160 [/latex]

We may have(1 red and 2 non-red) or (2 red and 1 non-red) or (3 red).
Required number of ways = [/latex](_{1}^{3}\textrm{C} \times _{2}^{9}\textrm{C}) + (_{2}^{3}\textrm{C} \times _{1}^{9}\textrm{C}) + (_{3}^{3}\textrm{C}) [/latex]
= (108 + 27 + 3)
= 138

A team of 6 players consisting of at least 3 girls can be formed in the following ways
Case I: When 3 girls and 3 boys are selected 3C6× 3C4
Case II: When 4 girls and 2 boys are selected 4C6 × 2C4
Case III: When 5 girls and 1 boy are selected 5C6 x 1C4
Case IV: When 6 girls and No boy are selected 6C6 x 0C4
Req. no. of selections = 6C3× 3C4 + 6C4 × 2C4 + 5C6 x 1C4 + 6C6 x 0C4
= 20 × 4 + 15 × 6 + 6 x 4 + 1 = 80+90+24+1 = 195