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Using only the digit from 0 to 5. How many 5 digit codes can be constructed? If the first digit cannot be 3 and number of digits cannot be used more than once.
Read Solution (Total 13)
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- 1st place can b filled by 1,2,4,5 ie, in 4 ways (3 and 0 can not b used)
and remainang 4 places can b filled by remaining 5 digits in 5P4 ways=120ways
so total 5 digit words =120*4=480 - 14 years agoHelpfull: Yes(21) No(2)
- 480
- 14 years agoHelpfull: Yes(8) No(0)
- 4*5*4*3*2=480
- 14 years agoHelpfull: Yes(5) No(0)
- 4ways_ 5ways__4ways_3ways_2ways=4*5*4*3*2=480
- 14 years agoHelpfull: Yes(5) No(1)
- its talking about 5 digit code not 5 digit no so the 1st digit may be 0
so the total posibilities is 5*5*4*3*2*1= 600
if that question for 5 digit no then the posibilities are 4*5*4*3*2*1=480 - 9 years agoHelpfull: Yes(3) No(0)
- 5c1*5c1*4c1*3c1*2c1=600
- 14 years agoHelpfull: Yes(2) No(11)
- 3*4*3*2*1=72
- 14 years agoHelpfull: Yes(2) No(5)
- total combination = 6p5=720
the number of combinations when 3 comes in first digit = 5p4 = 120
therefore answer will be 720-120= 600 - 14 years agoHelpfull: Yes(1) No(9)
- 72
- 14 years agoHelpfull: Yes(1) No(4)
- MR ROON I HAVE TO SAY U GUD AFTERNOON
first digit can't be zero so think it.. - 13 years agoHelpfull: Yes(0) No(2)
- (4)*5*4*3*2=480
- 9 years agoHelpfull: Yes(0) No(0)
- 3 cant be 1st digit
so..5*5*4*3*2 - 9 years agoHelpfull: Yes(0) No(0)
- 1st place= 3 ways, 2nd place= 4 ways, 3rd place=3 ways,4th place=2 ways,last=1
total by permutation = 3*4*3*2*1=72 - 8 years agoHelpfull: Yes(0) No(0)
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