how many integers can be formed using one or more of the digits 2,5,6,7, and 8 if no digit is repeated in a number?

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how many integers can be formed using one or more of the digits 2,5,6,7, and 8 if no digit is repeated in a number?

Mathematics
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infinite
fail^
for one digit there are 5 different numbers

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Other answers:

for 2digits there are ( 5 x 4 ) = 20 different numbers
for 3 digits there are ( 5 x 4 x 3 ) = 60 different numbers
for 4 digits there are 5 x 4 x 3 x 2 = 120 different numbers
for 5 digits there are 5 x 4 x 3 x 2 x 1 = 120 different numbers
total number of different numbers = 5 + 20 + 60 + 120 + 120 = 325
thank you so much for your help! I checked the packet answers and the answer is 195....? if it helps this is our probability unit
oh i see! i left out a very important word in the problem... it says *even positive* integers!
well thats means for one digit we have 3 different numbers
for two digits we have (3 x 4) = 12 different numbers for three digits we have ( 3 x 4 x 3 ) = 36 for four digits we have ( 3 x 4 x 3 x 2 ) = 72 for all five digits we have (3 x 4 x 3 x 2 x 1 ) = 72
add them and you get the 195
basically , you need to deal with the restriction first
thank you! how did you know to multiply it by 4 and 3 though?
for it too be even it must end in either 2, 6 or 8 ( three different ways )
so , take the 2digit case, first of all we must pick the last digit such that the number will be even ( we can do this in 3 ways ) , then once we have selected the last digit there are 4 digits remaining from which we can pick the second digit
similarly with the other cases
e.g. four digit case , first we pick the last digit ( which can be done in 3 ways ) then we have 4 digits remaining with which to choose the remaining 3 digits
oh! thank you soo much! I really appreciate it!

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