HELP please with this question!!?

Question

In how many ways can a group of 10 people be chosen from 6 adults and 8 children if the group must contain at least 2 adults?

Answer 1

There's no way in this case to avoid getting at least two adults (since we want to choose 10 people and there are only 8 children). So we can simply randomly choose the ten people, so there are 14 choose 10, or
14!/(10! * 4!) = 1001
ways to do this.

EDIT: A couple of answerers have fallen into the trap of trying to choose the two adults first, then pick out 8 more people from the remaining 12. This is very common, but it double-counts many of the combinations.

Let's use a smaller example to demonstrate this. Consider 3 adults and 1 child, and we want to choose 3 people, at least two adults. You can count by hand the number of possibilitites here, and should get 4 (we're leaving out one person, and each person left out is a different combination). However, the incorrect method tells us there are C(3,2)*C(2,1) = 6 ways to do it. So just what is it we're double counting? Think about what happens if we just choose all three adults. This method of counting actually counts it three times. We count it by choosing the adult A and adult B in the first step, and C in the second step; but also by choosing B and C in step 1 and A in step 2; and also by choosing A and C in step 1 and B in step 2.


So let's get back to our question: it's sort of easy, if you take a step away from it and realize that there's no way to get less than two adults. But you might want to know how to do this without the fancy sidestep. Let's say that there are 16 children instead to choose from.

The easiest way to do this now is to consider the number of adults, one by one, and to use the complement rule.

So the number of ways to choose 10 people with no adults is simply C(16,10). And the number of ways to choose exactly one adult is C(6,1)*C(16,9). [Note that this does not fall into the trap mentioned above, as the second choose function doesn't deal with any adults.] Finally, the number of ways to choose 10 people with no restrictions is C(22,10).

Therefore the number of ways to get at least 2 adults is
C(22,10) - C(16,10) - C(6,1)*C(16,9)
=569,998 ways (if my calculator worked it right).

(Note that if you try to use this method on the original problem, you'll get C(8,10) for the number of ways to choose zero adults, which is obviously a problem.)

Answer 2

Since there are 8 children, any group of 10 people must have at least 2 adults. So, the total number of possible groups is the total number of groups that meet the desired condition. Since we have 8 + 6 = 14 people and want 10 in each grooup, the answer is 14 choose 10, that is C(14, 10) = C(14, 4) = (14 * 13 * 12 * 11)/(4!) = 7 * 13 * 11 = 1001 groups. We have combinations because order is not important here.

Answer 3

Consider the 10 slots you are filling.
The first two must be adults, so the number of ways of filling slot 1 is 6 and the number of ways of filling slot two with the remaining adults is 5.
Now you have 8 remaining slots and 12 remaining people.
The next 8 slots will be filled as
12*11*10*9*8*7*6*5
Thus the total number of possible groupings (where order of selection matters) would be:
6*5*12*11*10*9*8*7*6*5
But the order you select the group doesn't matter, what matters is that they are in the group, not that they were picked first, second, etc.
Thus we need to divide by the total number of ways to choose 10 people: 10!
Tthis is the difference between Permutations (order matters, so there are more possibilities) and Combinations (order does not matter so there are fewer possibilities)

So your final answer is:
(6*5*12*11*10*9*8*7*6*5)/10!

Answer 4

C(6, 2)*C(12, 8)
where C(n, r)=n!/(r!*(n-r)!)

Answer 5

after 2 adults, people left = 8 + 6 - 2 = 12

first choose 2 adults out of 6adults, then 8 people from 12 people

so number of ways = (6C2)*(12C8) = 15*495 = 7425

Answer 6

Statistics NCR formula but I can not find that formula in the book. I am sure that is the one you use and then you just plug in the other numbers and into you calculator. Good luck I wish I still had my formula sheets The guy multiplying everything is correct but the formula is the easier way. It should be in your book

Answer 7

Yea, thats the equation