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This is a permutation problem.
The digits are: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
The 1st digit has 9 possibilities (it cannot be zero).
Thus, there are 9(9P9) = 9(9!) = 9(362,880) = 3,265,920 such numbers, because:
The 2nddigit has 9 possibilities (because we are left with 9 digits).
The 3rd digit has 8 possibilities.
The 4th digit has 7 possibilities.
The 5thdigit has 6 possibilities.
The 6thdigit has 5 possibilities.
The 7thdigit has 4 possibilities.
The 8thdigit has 3 possibilities.
The 9thdigit has 2 possibilities.
The 10thdigit has 1 possibility.
so that, 9*9*8*7*6*5*4*3*2*1= 3,265,920.
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How many 13 digit numbers are possible by using the digits 12345 which are divisible by 4 if repetition of digits is not allowed?
If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.
How many three digit numbers can be formed using the digits 873 and 6 without repetition?
I take it that you want to make three digits numbers with 8,7,3, and 6 without repetition. The first digit cane be selected from among 4 digits, the second from 3 digits, the third digit from 2, hence the number of three digit numbers that can be formed without repetition is 4 x 3 x 2 = 24
How many 3 digit numbers can be formed from the digits 1234?
64 if repetition is allowed.24 if repetition is not allowed.
Where d represents a digit from 0 9 how many different ISBN numbers are possible if repetition of digits is allowed?
There are 10 to the 10th power possibilities of ISBN numbers if d represents a digit from 0 to 9 and repetition of digits are allowed. That means there are 10,000,000,000 ISBN numbers possible.
How many 3-digit numbers can be formed by the digits 4 5 9 without repetition?
Six (6)
How many 13 digit numbers are possible by using the digits 12345 which are divisible by 4 if repetition of digits is not allowed?
If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.
How many 3 digit number can be made using the digits 2 4 7 9?
24 three digit numbers if repetition of digits is not allowed. 4P3 = 24.If repetition of digits is allowed then we have:For 3 repetitions, 4 three digit numbers.For 2 repetitions, 36 three digit numbers.So we have a total of 64 three digit numbers if repetition of digits is allowed.
How many three digit numbers can be formed using the digits 873 and 6 without repetition?
I take it that you want to make three digits numbers with 8,7,3, and 6 without repetition. The first digit cane be selected from among 4 digits, the second from 3 digits, the third digit from 2, hence the number of three digit numbers that can be formed without repetition is 4 x 3 x 2 = 24
How many seven digit numbers are there when 0 and 4 cannot be the leading number?
If the number can contain repeated digits, the answer is 800000. Without repetition, there are 483840.
How many 3 digit numbers are possible from the digits 012345?
Assuming that the first digit can't be zero: If repetition of digits is permitted: (5 x 6 x 6) = 180 numbers of 3 digits. If repetition of digits is not permitted: (5 x 5 x 4) = 100 numbers of 3 digits.
How many 3 digit numbers can be formed from the digits 1234?
64 if repetition is allowed.24 if repetition is not allowed.
How many 4 digit numbers can be formed from digits 123456789?
There are 7,290 different 4-digit numbers that can be formed from the digits 1-9 without repetition.
How many 4-digit numbers are possible if the hundreds digit is 8 and if repetition of digits is allowed?
-4
How many 4 digit numbers can be made to give the sum of 6?
If repetition of digits is allowed, then 56 can.If repetition of digits is not allowed, then only 18 can.
How many 2-digit numbers can be produce using the digits 123456789 if repetition is not allowed?
-123456787
Where d represents a digit from 0 9 how many different ISBN numbers are possible if repetition of digits is allowed?
There are 10 to the 10th power possibilities of ISBN numbers if d represents a digit from 0 to 9 and repetition of digits are allowed. That means there are 10,000,000,000 ISBN numbers possible.
How many 3-digit numbers using the digits 0-7 are possible if repetition are allowed?
290
