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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?
 2 years ago
 2 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?
 2 years ago
 2 years ago

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bahrom7893Best ResponseYou've already chosen the best response.1
1 digit  4 ways (any one can be chosen)
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.1
2  digits  4*3 ways (you picked the first one, so now you have 3 to choose from)
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.1
Waaait, isn't it 4C1
 2 years ago

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

adnanchowdhuryBest ResponseYou've already chosen the best response.0
Answer is supposed to be 64.
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.1
wait 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
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.1
ehhh... im lost.. crap i suck at probability
 2 years ago

adnanchowdhuryBest ResponseYou've already chosen the best response.0
Can 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'?
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.1
im 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
 2 years ago

adnanchowdhuryBest ResponseYou've already chosen the best response.0
So would one possibility be 1278?
 2 years ago

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

adnanchowdhuryBest ResponseYou've already chosen the best response.0
I understand now!
 2 years ago

adnanchowdhuryBest ResponseYou've already chosen the best response.0
4 possibilities for 1 digits. 4x3 = 12 possibilities for 2 digits. And so on..
 2 years ago

bahrom7893Best ResponseYou've already chosen the best response.1
yes that's what i was trying to say adnan
 2 years ago

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

amistre64Best ResponseYou've already chosen the best response.2
1, 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
 2 years ago

amistre64Best ResponseYou've already chosen the best response.2
lol, almost had that right
 2 years ago

amistre64Best ResponseYou've already chosen the best response.2
60+4 if i learn to count :)
 2 years ago

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

amistre64Best ResponseYou've already chosen the best response.2
not really; since it is split up into 4 groups to count individually you still have a bit of work
 2 years ago

amistre64Best ResponseYou've already chosen the best response.2
knowing the cPr stuff helps tho
 2 years ago

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

amistre64Best ResponseYou've already chosen the best response.2
yeah, 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
 2 years ago

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