TCS Permutations and Combinations 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)*(⁸C₁)


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

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


• Two digit no=4 * 5 = 20


• Three digit no=4 * 5 * 5 = 100
  
  
  
  • Four digit no=3 * 5 * 5 * 5 = 375(the possibility for 1,2,3 will come in the first position)
  
  
  • Four digit no=1 * 3 * 5 * 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*4*3*2*2*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

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!*18..


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

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


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


• Ways of forming two digits without repetition=10*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)