InfyTQ Permutation and Combination Quiz 1

a,b,c,d are grouped ie consider them as one and remaining as 6.
total 6+1=7! ways

To form a triangle we need 3 points

• Select 2 points from the 10 points of line AB & 1 from the 8 on AC = (¹⁰C₂) x (⁸C₁)


• Select 2 points from the 8 points of line AC & 1 from the 10 on AB= (⁸C₂) x (¹⁰C₁₎

NO OF 4'S NO. of 6's
50 0
47 2
44 4
41 6
38 8
35 10
32 12
29 14
26 16
23 18
20 20
17 22
14 24
11 26
8 28
5 30
2 32

So total 17 ways but here it is 4's & 6's both so don't consider 1st one
Final answer: 16 ways

4x + 6y = 200

⇒2x+3y=100

⇒x=\frac{(100−3y)}{2}=50−(\frac{3}{2})y
As x is an integer, y should be a multiple of 2.
If
y = 0, x = 50
y = 2, x = 47
y = 4, x = 44
...
y = 32, x = 2

So total ways are \frac{(32-0)}{2} + 1 = 17 ( if 0 6s are possible) otherwise 16

Ans is 575

• One digit no = 4 (0 is not a positive integer)


• Two digit no=4 x 5 = 20


• Three digit no=4  x 5 x  5 = 100
  
  
  
  • Four digit no=3 x 5 x 5 x 5 = 375(the possibility for 1,2,3 will come in the first position)
  
  
  • Four digit no=1 x 3 x 5 x 5 (the possibility of 4 is fixed in the first position and then 0,1,2 is comes in second position)
  
  
  
  


• And the last digit is 4300 we include this number also

given 6th digit even number , so last digit 2 or 4 or 6-> 3 ways
" 5th digit should be even...so there will be 2 ways(rep. not allowed)
so,therefore we get 5 x 4 x 3 x 2 x 2 x 3=720 ways

Single digit number plates =   ⁵C₁ x 1!
Two digit number plates =   ⁵C₂ x 2!
Three digit number plates =   ⁵C₃ x 3!
Four digit number plates =   ⁵C₄ x 4!
Five digit number plates =   ⁵C₅ x 5!
Total = (⁵C₁ x 1!) + (⁵C₂ x 2!) + (⁵C₃ x 3!) + (⁵C₄ x 4!) + (⁵C₅ x 5!) = 5 + 20 + 60 + 120 + 120 = 325

18! x 2

References :

1. Circular Permutations Formula


2. How to solve Circular Permutations


3. Tips and Tricks to solve Circular Permutations

• Consider 1 man along with two sisters as one group.. so 18 objects, they can be arranged in 17! ways as circular. (Formula for Circular Permutation is : (n - 1)!


• The one man in between the two sisters can be out of any 18 men so, 17! x 18..


• The two sisters can be arranged in 2 ways so 18! x 2

• No.of alphabets=26 (a-z), no.of digits=10(0-9).


• Ways of arranging two alphabets with out repetition=26 x 25;


• Ways of forming two digits without repetition=10 x 9

Following would possible scenarios: Men - (0, 1, 2, 3)

If there are 0 men, then 11 women. Similarly, if there are 1 men then there would be 10 women so on ...

• Ways of choosing (0 Men, 11 Women) : (⁵C₀ × ¹¹C₁₁)


• Ways of choosing (1 Men, 10 Women) : (⁵C₁ × ¹¹C₁₀)


• Ways of choosing (2 Men, 9 Women) : (⁵C₂ × ¹¹C₉)


• Ways of choosing (3 Men, 8 Women) : (⁵C₃ × ¹¹C₈)