total no.of 4digit no.having all diff digit is equal to??
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Answered by
2
6
down vote
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Ok so lets write down any old 4 digit number
abcd
How many choices do we have for the digit a? We have 9 choices (since the first digit cannot be 0). Now for each possible choice of a we have 9 choices for b (since we want b to be a different digit to a and we now allow 0).
So for choice of the ab part we have 9∗9=81 possibilities.
Now for each of these we have 8 choices for c (to avoid c being the same as either a or b). And for each of these we have 7 choices for d (to avoid d being the same as either a,b or c).
So in total there are 9∗9∗8∗7=4536 possible numbers.
down vote
accepted
Ok so lets write down any old 4 digit number
abcd
How many choices do we have for the digit a? We have 9 choices (since the first digit cannot be 0). Now for each possible choice of a we have 9 choices for b (since we want b to be a different digit to a and we now allow 0).
So for choice of the ab part we have 9∗9=81 possibilities.
Now for each of these we have 8 choices for c (to avoid c being the same as either a or b). And for each of these we have 7 choices for d (to avoid d being the same as either a,b or c).
So in total there are 9∗9∗8∗7=4536 possible numbers.
Answered by
1
For a valid 44-digit number, the Highest Significant digit will be among 1⋯91⋯9 so can have 99 values
From the next onwards, they can assume 0⋯90⋯9 excluding the ones already selected
So, the number of combinations will be
9⋅9⋅8⋅7
Hint:
The first digit can be any number from 11to 99. The next three digits can be any number from 00 to 99. If we choose each digit one at a time and stipulate that each digit is different from the previous choices, how many choices do we have for each digit? Multiply these together to get your answer.
Allowing for a zero first digit
10 * 9 * 8 * 7
10 possible options for the first digit, 9 remaining options for the second digit, 8 remaining options for the third digit
From the next onwards, they can assume 0⋯90⋯9 excluding the ones already selected
So, the number of combinations will be
9⋅9⋅8⋅7
Hint:
The first digit can be any number from 11to 99. The next three digits can be any number from 00 to 99. If we choose each digit one at a time and stipulate that each digit is different from the previous choices, how many choices do we have for each digit? Multiply these together to get your answer.
Allowing for a zero first digit
10 * 9 * 8 * 7
10 possible options for the first digit, 9 remaining options for the second digit, 8 remaining options for the third digit
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