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 3 years ago
How many different numbers can be formed by taking one, two, three and four digits from the digits 1, 2, 7 and 8, if repetition are not allowed?
 3 years ago
How many different numbers can be formed by taking one, two, three and four digits from the digits 1, 2, 7 and 8, if repetition are not allowed?

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bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.11 digit  4 ways (any one can be chosen)

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.12  digits  4*3 ways (you picked the first one, so now you have 3 to choose from)

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1Waaait, isn't it 4C1

adnanchowdhury
 3 years ago
Best ResponseYou've already chosen the best response.0I don't understand. Can you list the permutations? I can't seem to understand what the question is saying.

adnanchowdhury
 3 years ago
Best ResponseYou've already chosen the best response.0Answer is supposed to be 64.

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1wait im wrong.. hold up. You have to use that formula, like the number of ways of choosing some object from a set of objects with no repetitions

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1ehhh... im lost.. crap i suck at probability

adnanchowdhury
 3 years ago
Best ResponseYou've already chosen the best response.0Can you explain what the question is asking? What does it mean when it says: 'take one, two, three and four digits from the digits 1, 2, 7 and 8'?

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1im thinking: Pick one digit from 1,2,7 and 8. Like I can pick 1, or 2, or 7, or 8, so the answer's 4 ways

adnanchowdhury
 3 years ago
Best ResponseYou've already chosen the best response.0So would one possibility be 1278?

adnanchowdhury
 3 years ago
Best ResponseYou've already chosen the best response.0One digit possibilities: 1, 2, 7 and 8. So there are 4 permutations for one digits...?

adnanchowdhury
 3 years ago
Best ResponseYou've already chosen the best response.0I understand now!

adnanchowdhury
 3 years ago
Best ResponseYou've already chosen the best response.04 possibilities for 1 digits. 4x3 = 12 possibilities for 2 digits. And so on..

bahrom7893
 3 years ago
Best ResponseYou've already chosen the best response.1yes that's what i was trying to say adnan

adnanchowdhury
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks! My problem was in understanding what the question was asking, but after you explained it, i understood.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.21, 2, 7 and 8 1 digit numbers; there are only 4 numbers so: 4 2 digit number: 12 17 18; 4 times = 12; 4P1 * 3P1 = 12 3digit number: 127 128 172 178 182 187; 6*4 = 24; 4*3*2 = 24 4 digit numbers: 1278 1287 1728 1782 1827 1872: 6*4 = 24; 4*3*2*1 = 24 4+12+24+24 = 4+12(3) = 40 i beleive

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2lol, almost had that right

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.260+4 if i learn to count :)

adnanchowdhury
 3 years ago
Best ResponseYou've already chosen the best response.0Any formula i could have used to save me time?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2not really; since it is split up into 4 groups to count individually you still have a bit of work

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2knowing the cPr stuff helps tho

adnanchowdhury
 3 years ago
Best ResponseYou've already chosen the best response.0I understood the formula before. But now, I just find it confusing.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.2yeah, its just a "structured" way to count and if you dont use it you tend to go back to the basic method of 1,2,3,4,5s

adnanchowdhury
 3 years ago
Best ResponseYou've already chosen the best response.0I see. I work it out intuitively now  probably takes me a bit longer, but i get there eventually.
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