CPhill

(i)  We need to have a common basis for the ratios

Men : Children  =   11 : 2   =  33 : 6

Woman : Children  = 10 : 3  =  20 : 6

Therefore  .......  Men : Women  =  33 : 20

(ii) The number of men  =   (33/6) * 24  =  132

     The number of women  = (20/6) *24  = 80

     Total members  =   132 + 80 + 24  =  236

Here's a lovely pair.....so stylish...............

Ah .....yes, GA.....I missed that.....!!!

I guess I'd better leave before someone tells me, "Looks like your CDD is flaring up again."

The "O's" at the beginning and end are indistinguishable

So.....we have 4!  = 24 possible arrangements  of the middle letters "PTAT'

But the "T's"  are also indistinguishable....so we need to divide by  2  = 12 distinguishable words....

"At most" means either   no dice show a "6",  1 die shows a "6",  2 die show a "6"  or  3 die show a "6"

No die shows a "6'  = (5/6)^4  = 625/1296

I die shows a "6"  =   C(4,1) * (1/6)*(5/6)^3  = 125/324  = 500/1296

2 die show a "6"  = C(4,2) * (1/6)^2*(5/6)^2  =  25 /216  = 150 / 1296

3 die show a "6"  =  C(4,3) *(1/6)^3 * (5/6)  = 5/324  =  20/ 1296

P ( getting at most three 6's on top)  = [ 625 + 500 + 150 + 20 ] / 1296  =

0.9992283950617284  ≈   99.99%   [almost a sure thing....!!!!]

I'll give this a shot.....

We have  C(6,2)  = 15 possible  pairs of answers

For 1/2 credit,  each of the correct  answers will be paired 4 times with incorrect answers  = 8 pairs

So ....  P(1/2 credit)  =  8 / 15   ≈  53.3%

Assuming that we could have a number with all 0's, there are 100,000 numbers possible  [10 digits possible in each of the 5 places = 10^5 = 100,000  ]

We want to choose the sets of any  5 of the 10 numbers permuted  = P(10,5)  = 30,240

P = 30,240 / 100,000  =  30.24%

There are  3!  = 6 possible orders.....we are interested in either   { S, C ,P}  or {C, S, P}

So .....  2/6  = 1/3  ≈  33.33%

Nice, hectictar.....!!!!