A census taker visits some number of families and determines that on average they have 2.6 children.

(a) What is the least number of families in this census?

(b) Given that the least number of families was polled, how many children in total did the families have?

(c) So, (a) and (b) are too easy? Agreed. So, in how many ways may these children have been distributed among the families (e.g., one family could have had all of them, or they could have been split among two families, etc.)?

Answers from a non-actuary: (a) 5; (b) 13; and (c) a huge number!

You done good and in New Math (Math, Nu?) circles, we award an "A" for recognizing that though likely only about one sixth billionth of the US Natioinal Debt, (c) is a huge number

If it's only a six billionth of the national debt, (c) is not as big a number as I'd figured it would be . . .

Let x = 1, then the answer is:

((x+1) x 10^(x+1) + (x+2) x 10^x + (x+3)² x 10⁰ / 2) x 10¹

Right?

We are talking about arranging 13 indifferent objects [kids] among 5 boxes [families] with blanks [no pun intended] permitted. Then, evaluate _[n+r-1]C_[r-1]:

_[13+5-1]C_[5-1]=₁₇C₄=2,380

Best to test the formula with small numbers, like 5 objects among 3 boxes = _[5+3-1]C_[3-1]=₇C₂=21. It's fairly simple to write out the cases and see that you get 21.

Did you evaluate my expression?

No, because I can't see your face.

As to evaluating the math, I have to count on my fingers, (but I can guess which one you are holding up right now) and even then I get mixed up, so I'm in complete awe of those of you to whom this stuff makes sense.

I get something like (1 to the hob) plus (2 to the fob) which gives you (3 to the hobenfoben) divided by the (fobenhoben) times the (hobfobhobentofobenfobenhoben) which of course equals the hobfob which everyone knows is the BoSox team ERA for next year.

Let x = 1, then the answer is:

((x+1) x 10(x+1) + (x+2) x 10x + (x+3)2 x 100 / 2) x 101

Right?

x=1 ==> (2 x 10^2 + 3 x 10 ^ 1 +16 x 1/2] x 10 = (200 + 30 + 8) x 10 = 2,380

Yes, I had, but was to0 dense to see if your expression had anything to do with how you solved the problem or you solved the problem in some other fashion and then you simply wanted to be cute.

Instead of _xC<sub>y I've always been fond of:

( ^x )
^y

But darn if I know how to format that so that it looks right. You didn't ask for the underlying formulas.</sub>

A got a Valentine's Day card once that was signed XXOOXXOOXXOO. Does that mean my secret admirer was an actuary? Or is that, perhaps, the formula that solves Andy's hypothetical lazy census taker question?

In the ancient books I read, n! is represented as | ⁿ

Sieve-

I thought that meant you were considered to be an OX-e- Moron

Tom -- Perhaps you speak the truth. Of course, if given the choice, 'tis better to mess with a hamster-e-moron than an ox-e-moron.

Perhaps a football coach?

For sake of illustration, assume 6 (n items) identical items, x, going into 4 (r boxes) boxes. Let's assume no vacant boxes for the time being.

Then, we look at the number of intervals as x|X|X|X|X|X. So, there are 5 intervals, or in the general sense, n-1 intervals.

Now, let's just look at the objects partitioned into one arrangement into the four boxes: xxx|x|x|x. Note, there are three partitions, or in the general sense r-1 partitions. Then if we look at the number of combinations of n-1 intervals taken r-1 at a time, we have the answer when blank boxes are not permitted. This is _n-1C_r-1. To see this it may be helpful in the above arrangements to blank out the x's in your mind and think of the intervals in the second paragraph as mail box pidgeon holes and the partitions as marbles.

Now, suppose blank boxes are permissible. Then, rather than looking at n items, we're really looking at n+r items where we will distributing r blank items as if they were real items. So, the answer is _n+r-1C_r-1