Consider all permutations (i.e., arrangements) of digits 1, 2&3. We will say that a hit has been scored if at least one digit occurs in its proper position in the permutation. If 1 (one) occurs in the first position or 2 in the second position or 3(three) in the 3rd position in how many ways of these permutations is a hit scored?
a) 4 b) 1 c) 2 d) 3

• 123,132,321,213.......so, ans is a)4
• 12 years agoHelpfull: Yes(42) No(2)
• 1 can be arrange in 123, 132
  2 can be arrange in 321
  3 can be arrange in 213
  
  number of possobolity is 4.
• 12 years agoHelpfull: Yes(18) No(0)
• ans. is 4 becs we say a hit whan it is in correct number position means 1 in first place.2 in second place and so on;so we say the hit in following condition.
  1. 123
  2. 132
  3. 213
  4. 321
• 12 years agoHelpfull: Yes(6) No(1)
• 123
  132
  321
  213
  .......so, ans is
  a)4
• 12 years agoHelpfull: Yes(3) No(0)
• total no of permutations will be 1 in which there will be exactly one way of representing 123.
• 12 years agoHelpfull: Yes(1) No(6)
• what is the correct answer???? i think it should be 3.
• 12 years agoHelpfull: Yes(0) No(8)
• answer is 4
  totally there can be 6 permutations
  1)123
  2)132
  3)213
  4)231
  5)312
  6)321
  out of which 1,2,3,6 satisfies the condition..
  so ans:4
• 11 years agoHelpfull: Yes(0) No(0)