Can you determine the possibility of choosing 6 items from 10 ones?

Answer 1

I think you meant the "PROBABILITY"? --- although even that isn't what you really had in mind. [The POSSIBILITY is presumably 1, since it's always POSSIBLE. However, if this was simply a trick question, I claim to have been the first person to answer it!]

I presume that you really want to know the NUMBER OF WAYS that you can pick different groups of 6 items from among a total of 10 items. Although the techniques used for this are also frequently used in Probability Theory (and this question may have arisen in a course on Probability Theory), the answer to this particular question is itself NOT a "probabilty" --- it's simply the number of ways that it can be done, without any regard to, or direct connection with, "probability" in this particular context.

In any case, back to the NUMBER OF WAYS that this can be done:

It's not "210!" but rather: 210.

[It is dangerous, or at least possibly misleading, in a situation where the "factorial symbol" (" ! ") appears so much, to put a " ! " immediately after any number if you DON'T REALLY MEAN "factorial."

The general formula for picking out r objects from a total number of n objects, WITHOUT REGARD TO ORDER OF SELECTION, is:

n! / [r! (n-r)!].

In the present case, that gives 10! / [6! 4!] = 10*9*8*7/[1*2*3*4] =
10*3*7 = 210. QED.

Here's how you establish the general result quoted:

Array the n objects before you. You can select the 1st object in n ways, then the 2nd object [from the remaining (n - 1) objects] in (n - 1) ways, the 3rd in (n - 2) ways, ... the rth (and last) object (n - r + 1) ways. (Just keep following the mathematical pattern established by considering selecting the first few of them.)

The number of ways that you can have collected the r objects, so far, is:

n (n - 1) (n - 2) ... (n - r + 1) = n! / (n - r)!

However, any one of the sets of r objects present in this number just written could have itself been chosen in r! ways, the number of differently ordered permutations of those r objects.

For example, suppose r were 3. You could have chosen them in the 6 different orders:

123, 132, 213, 231, 312, 321 --- and that's all, 3! or 6 ways.

That means that the first way of selecting groups of objects overcounted by a factor of r! --- as originally selected, any group of r would have appeared in r! ways of permuting its members' individual ordering position.

So, the number of DIFFERENT GROUPS of r objects that you can select from n objects, WITHOUT REGARD TO THEIR ORDER OF SELECTION, is, as stated near the outset:

n! / [r! (n-r)!].

Hence the conclusion reached for this particular application: 210.

Live long and prosper.

P.S. Oh, dear, you may have had in mind yet another variant of the question you actually posed :

"If I choose 6 items from 10 items (not all "ones"!),*** what is the PROBABILITY that they (the 6 chosen) will match some predetermined set of 6 previously chosen at random by myself or someone else?"

Now THAT is a question in which PROBABILITY properly arises. In that case, since only one set can be correct, yet there are 210 ways in which sets of 6 can be chosen, the PROBABILITY of matching the target set is the FRACTION:

1/210.

*** Note: if they are all "ones" (you said in your question that there were "10 ones" !), then again, the probability of matching the original selection is precisely 1 since the only possibility is:

1, 1, 1, 1, 1, and 1,

and any 6 out of 10 "ones" will match perfectly!

I hope that I've convinced you that, at the very least, you should do your very best to express your question clearly, so as not to send those trying to serve your interests off on various wild goose chases.

Answer 2

Choosing 6 items out of 10 regardless the order of choosing = 10C6 which gives you 210.

When first item is chosen, you are given 10 choices. Next, 9, and then followed by 8, 7, 6, and finally 5 as the 6th item u pick.

Hence, probability = (10C6) / (10*9*8*7*6*5) = 210 / 151200 = 1/720

Answer 3

If I understand your question, you want the chances of picking six particular items out of a group of 10, presumably drawing one at a time, at random.

For the first pick, you have 6 "right" choices, out of a possible 10. For the second, 5 out of 9; for the third, 4 out of 8 and so forth. The total then comes to:
6/10 * 5/9 * 4/8 * 3/7 * 2/6 * 1/5, or 720/151200 or 1/210.

Not very likely. Hope it helps.

Answer 4

it's easy

the first time you pick, there are 10 items that you could possibly get...

10

the second time, you only have 9 to choose from...

10 * 9

the third time, there are only 8 objects left...

10 * 9 * 8

and so on, until you've picked 6 times

10 * 9 * 8 * 7 * 6 * 5

then you multiply it all to get 151200!

i think that was what the question is asking..if not, post additional info to clarify :)

Answer 5

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Answer 6

210!