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if the repetition is not allowed then the arrangement can be 3*2*1 = 6 ways

if repetition is allowed the ways can be 6*6*6= 216 ways

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Q: What are the possible variations of a 5 digit sequence using 1-6 when 345 are the first 3 digits?

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if repeating is allowed... 36 (6x6, for the last two digits) If not, 6 (3x2, last two digits)

3!=6

If repeated digits are allowed, then the largest 12-digit number is 999,999,999,999 .If repeated digits are not allowed, then the largest 12-digit number is 989,898,989,898 .If the same digit can't be used more than once, then the largest possible number has only10 digits. The number is 9,876,543,210 .

There are 2 possible digits for the first digit (3 or 4), leaving 3 possible digits for the second digit (5 and 6 and whichever was not chosen for the first), leaving 2 possible digits for the third. Thus there are 2 × 3 × 2 = 12 possible 3 digit numbers.

There are 30,240 different 5-digit numbers. Math: 10*9*8*7*6 1st digit has 10 possible choices (0-9) 2nd digit has 9 possible choices (one of the digits was used in the 1st digit) 3rd digit has 8 possible choices 4th digit has 7 possible choices 5th digit has 6 possible choices

There are two possible digits for the first and last digit, and two possible digits for the centre digit, making 2 × 2 = 4 possible 3 digit palindromes from the set {1, 2}, namely the set {111, 121, 212, 222}.

Assuming the the last part of the question is that the sequence may not END in 000, there are 2997 sequences.

There are seven possible digits for the first digit and 6 digits for the second (minus one digit for the digit used as the first digit) and 5 options for the last digit (minus one again for the second digit) and then you just multiply them all together to get a total possible combination of 210 numbers that are possible.

what is the greets possible 9 digit number that uses each of the digits 1-3 times

If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.

The answer depends on which digit is 5 and what the other digits are!

Assuming that 2356 is a different number to 2365, then: 1st digit can be one of four digits (2356) For each of these 4 first digits, there are 3 of those digits, plus the zero, meaning 4 possible digits for the 2nd digit For each of those first two digits, there is a choice of 3 digits for the 3rd digit For each of those first 3 digits, there is a choice of 2 digits for the 4tj digit. Thus there are 4 x 4 x 3 x 2 = 96 different possible 4 digit numbers that do not stat with 0 FM the digits 02356.

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