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1 digit - 4 ways (any one can be chosen)
2 - digits - 4*3 ways (you picked the first one, so now you have 3 to choose from)
Waaait, isn't it 4C1
I don't understand. Can you list the permutations? I can't seem to understand what the question is saying.
Answer is supposed to be 64.
For which one?
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
ehhh... im lost.. crap i suck at probability
Let me call amistre
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'?
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
yay amistre's here
So would one possibility be 1278?
One digit possibilities: 1, 2, 7 and 8. So there are 4 permutations for one digits...?
I understand now!
4 possibilities for 1 digits. 4x3 = 12 possibilities for 2 digits. And so on..
yes that's what i was trying to say adnan
Thanks! My problem was in understanding what the question was asking, but after you explained it, i understood.
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
lol, almost had that right
60+4 if i learn to count :)
Any formula i could have used to save me time?
not really; since it is split up into 4 groups to count individually you still have a bit of work
knowing the cPr stuff helps tho
nPr ... cant type
I understood the formula before. But now, I just find it confusing.
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
I see. I work it out intuitively now - probably takes me a bit longer, but i get there eventually.