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Mathematics 20 Online
OpenStudy (anonymous):

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?

OpenStudy (bahrom7893):

1 digit - 4 ways (any one can be chosen)

OpenStudy (bahrom7893):

2 - digits - 4*3 ways (you picked the first one, so now you have 3 to choose from)

OpenStudy (bahrom7893):

Waaait, isn't it 4C1

OpenStudy (anonymous):

I don't understand. Can you list the permutations? I can't seem to understand what the question is saying.

OpenStudy (anonymous):

Answer is supposed to be 64.

OpenStudy (bahrom7893):

For which one?

OpenStudy (bahrom7893):

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

OpenStudy (bahrom7893):

ehhh... im lost.. crap i suck at probability

OpenStudy (bahrom7893):

Let me call amistre

OpenStudy (anonymous):

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'?

OpenStudy (bahrom7893):

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

OpenStudy (bahrom7893):

yay amistre's here

OpenStudy (anonymous):

So would one possibility be 1278?

OpenStudy (anonymous):

One digit possibilities: 1, 2, 7 and 8. So there are 4 permutations for one digits...?

OpenStudy (anonymous):

I understand now!

OpenStudy (anonymous):

4 possibilities for 1 digits. 4x3 = 12 possibilities for 2 digits. And so on..

OpenStudy (bahrom7893):

yes that's what i was trying to say adnan

OpenStudy (anonymous):

Thanks! My problem was in understanding what the question was asking, but after you explained it, i understood.

OpenStudy (amistre64):

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

OpenStudy (amistre64):

lol, almost had that right

OpenStudy (amistre64):

60+4 if i learn to count :)

OpenStudy (anonymous):

yes lol

OpenStudy (anonymous):

Any formula i could have used to save me time?

OpenStudy (amistre64):

not really; since it is split up into 4 groups to count individually you still have a bit of work

OpenStudy (amistre64):

knowing the cPr stuff helps tho

OpenStudy (amistre64):

nPr ... cant type

OpenStudy (anonymous):

I understood the formula before. But now, I just find it confusing.

OpenStudy (amistre64):

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

OpenStudy (anonymous):

I see. I work it out intuitively now - probably takes me a bit longer, but i get there eventually.

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