@joe-shmo said
I believe 4 Saints have Halos. If I'm right I'll go through the solution.

Correct.Well done.

@venda said
Correct.Well done.

Ok!

The person making the puzzles must be on a combinatorics kick. So the objective is to divvy up the 7 halos among the 7 saints leaving a certain number of saints without halos.

The solution uses the formula I derived for the last problem you posed about 4 people splitting 11 chickens. That was convenient!

Redefine "x" to be the number of saints without halos.

lets start with x = 0 ( every saint has a halo ) I know this isn't possible from the problem statement but it helps describe the progression.

Dividing up 7 halos among 7 saints gives 1 icon.

1 + 1 + 1 + 1 + 1 + 1 + 1
◙....◙....◙....◙....◙....◙....◙

The sum represents the number of halos on each saints head and the ◙ are the saints with halos and the ■ are the saints without ( this assumes the saints are indistinguishable from each other otherwise there would be 7! icons in this setup alone )

So now we go to x= 1 ( 1 saint without halo )

We break the 7 halos up into 6 summands

1○1○1○1○1○1○1

I showed how to do this in the other chicken problem.

Let S = 7 be the number of halos ( represented by the string of 7 1's above )
Let N be the number of summands ( i.e how many of the saints have halos )

If we choose 5 ○'s ( out of the 6 ○'s in total ) to be + signs we will get one instance of 7 halos shared amongst the 6 saints with halos. For this particular case we see that is equivalent to choosing which circle is NOT a "+ sign" and it should become obvious there are 6 ways of doing this.

1○1+1+1+1+1+1 = 2+1+1+1+1+1
1+1○1+1+1+1+1 = 1+2+1+1+1+1
1+1+1○1+1+1+1 = 1+1+2+1+1+1
1+1+1+1○1+1+1 = 1+1+1+2+1+1
1+1+1+1+1○1+1 = 1+1+1+1+2+1
1+1+1+1+1+1○1 = 1+1+1+1+1+2

Now lets take the 6 sum's and apply them to the to the saints; saying the farthest saint to the left is without a halo:

0 + 2 +1 +1 + 1 +1 +1
■...◙...◙...◙...◙...◙...◙

0 + 1 +2 +1 + 1 +1 +1
■...◙...◙...◙...◙...◙...◙

0 + 1 +1 +2 + 1 +1 +1
■...◙...◙...◙...◙...◙...◙

0 + 1 +1 +1 +2 +1 +1
■...◙...◙...◙...◙...◙...◙

0 + 1 +1 +1 +1 +2 +1
■...◙...◙...◙...◙...◙...◙

0 + 1 +1 +1 + 1 +1 +2
■...◙...◙...◙...◙...◙...◙

However, assigning the left most saint to be the one without a halo was completely arbitrary. It could be any of the saints. There are 7 ways to choose which saint does not have a halo.

■...◙...◙...◙...◙...◙...◙
◙...■...◙...◙...◙...◙...◙
◙...◙...■...◙...◙...◙...◙
◙...◙...◙...■...◙...◙...◙
◙...◙...◙...◙...■...◙...◙
◙...◙...◙...◙...◙...■...◙
◙...◙...◙...◙...◙...◙...■

Now we have our two relationships.

The number of distinct icons that can be made is the number of ways to split 7 halos amongst ( 7 - x ) saints times the number of ways to choose x saints of 7 saints to NOT have halos.

Let I(x) be the number of distinct icons

I(x) = [ 6!/( (6-x)!*x! ) ]*[ 7!/( x!*(7-x)! ]

From here ( using a spreadsheet ) check the values of x.

We see at x = 3 ( 3 saints without halos )

I ( 3 ) = [ 5*2*2 ]*[7*5] = 700

And we are done!

Another thing to note that the problem writers stressing that exactly "x" saints have halos ( "7-x" saints as I've defined x ) is important. If they allow the number of saints that have halos vary ( some icons may have 2,3,6 halos etc...) you can sum them up in a particular way to get exactly 700 icons as well.

@venda said
I approached it in a slightly different way:-
700 is always the key in these things(I know that from reading the paper every week!)
So the multiples of the number of combinations of saints v icons(7!/x!) multiplied by the possible ways of dividing the halos between the 7 saints as in the chicken problem(7!-1/x!-1))must equal 700
I discounted 1 and 7 (obviously).700 does not ...[text shortened]...
plugging 4 into the equations yields 35*20 which equals 700 so the answer is 4 saints(x) have halos

Something is wrong with the formulas you've written out, but those are the same results...so you 've probably just misrepresented them a bit. Also, ours are basically the same method. Good solve!

@joe-shmo said
Back to the Bezique problem.

The probability of holding double Bezique as I understand it ( 2, Q, S+ 2, J, D ) after drawing 32 cards ( half of the deck - your opponent drawing the other half ) is:

C ( 32,4 )*C( 4,2)*2*2/( 64*63*62*61 ) = 35960*6*2*2/( 64*63*62*61 ) ≈ 6/100

Thanks for all your input my friend.
I'll keep looking at the Times puzzles to see if anything interesting comes up.