a security code for a building consist of 6 numbers what is the number of security codes possible if the first number can not be a 9
900,000 ways
how did you get that ans.
is that 1x2x3x4x5x6 kind of problem
9x10^5 since the first number is not 9 the choices are form 0 to 8 digit so you have 9 choices then for the next digits you have 10 options per digit that why you will come up with 9x10x10x10x10x10 =9x10^5 = 900,000 ways.
Thank you for all your help
9x10x10x10x10x10 1. 9 digits can go on 1st place (0,1,2,3,4,5,6,7,8) 9 possible combinations 2. 10 digits can go on 2nd place (0,1,2,3,4,5,6,7,8,9) 10 possible combinations 3. 10 digits can go on 3td place (0,1,2,3,4,5,6,7,8,9) 10 possible combinations 4. 10 digits can go on 4th place (0,1,2,3,4,5,6,7,8,9) 10 possible combinations 5. 10 digits can go on 5th place (0,1,2,3,4,5,6,7,8,9) 10 possible combinations 6. 10 digits can go on 6th place (0,1,2,3,4,5,6,7,8,9) 10 possible combinations You multiple all possible combinations to get the total number of combinations. The result will be different if you are not allowed to replace the digit after it is picked. 1. 9 digits can go on 1st place (0,1,2,3,4,5,6,7,8) 9 possible combinations 2. 9 digits can go on 2nd place (0,1,2,3,4,5,6,7,8,9 minus the picked from previous step(s)) 9 possible combinations 3. 8 digits can go on 3td place (0,1,2,3,4,5,6,7,8,9 minus the picked from previous step(s)) 8 possible combinations 4. 7 digits can go on 4th place (0,1,2,3,4,5,6,7,8,9 minus the picked from previous step(s)) 7 possible combinations 5. 6 digits can go on 5th place (0,1,2,3,4,5,6,7,8,9 minus the picked from previous step(s)) 6 possible combinations 6. 5 digits can go on 2nd place (0,1,2,3,4,5,6,7,8,9 minus the picked from previous step(s)) 5 possible combinations tottal = 9x9x8x7x6x5 = 136080
if the restriction of the digit 9 not being allowed in 1st place is removed you will have 10x10x10x10x10x10 = 1000000 (with replacement) 10*9*8*7*6*5 = 151200 (with no replacement)
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