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1.The digits 1234 are placed in 3*3 matrix such that no same number are in same row and column. how many many such ways are possible???
Read Solution (Total 12)
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- positions are
1 2 3
4 5 6
7 8 9
ways of filling each place raw wise keeping the conditions in mind can b
4c1 3c1 2c1
3c1 2c1 1c1
2c1 1c1 1c1
ans =4*3*2*3*2*1*2*1*1=288
- 11 years agoHelpfull: Yes(22) No(0)
- way of filling the 3*3 matrix will be
4 3 2
3 2 2
2 2 2
so, 4*3*2*3*2*2*2*2*2 - 11 years agoHelpfull: Yes(16) No(16)
- 4*3*2*3*2*2*2*1*1
- 11 years agoHelpfull: Yes(5) No(7)
- plz explain
- 11 years agoHelpfull: Yes(3) No(0)
- @ Vishakha the product of each row has to be added or multiplied plz reply
- 11 years agoHelpfull: Yes(2) No(0)
- @ Annu tera solution sahi hai na
- 11 years agoHelpfull: Yes(2) No(1)
- total possibility 3*3=9
no of arrangment (1,2,3)
(4,1,2)
(3,4,1)
so there are 3 possibility so,3/9=1/3 ways,,, - 11 years agoHelpfull: Yes(1) No(10)
- it should have 5 digits with four digits there comes a position where the column above it contains 2 nos and its row contains two nos then their wont be a number left.
- 11 years agoHelpfull: Yes(1) No(0)
- the ans is 44
- 11 years agoHelpfull: Yes(0) No(10)
- plzz clearly explain any one
- 11 years agoHelpfull: Yes(0) No(0)
- it is possible to create 4*4 matrix such that no same number are in same row and column with 1234.so for each row and column it will be 4c3*4c3=16
- 11 years agoHelpfull: Yes(0) No(0)
- @sachdev we can fill the 2nd row 3rd column in 2c1 ways since u selected 2nd row 1st colum in 3 ways and the other 1 left can be used in 3rd row 2nd column and same with 3rd row 2nd column
- 11 years agoHelpfull: Yes(0) No(0)
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