September 17th, 2013

turtle

Miscellaneous skills they don't teach in seminary #2,479

We have a very nice flagpole in the back garden of the church. Its platform and landscaping were an Eagle Scout service project some years ago. The Scouts use it for ceremonies and to teach proper handling of the flag.

I received word that we were supposed to lower the flag to half mast through Friday, by Presidential order re: the Navy Yard shooting. So I went out to the flagpole last night to lower the flag and discovered it was stuck. Whoever had last raised it not only left a rat's nest of a knot on the pole, but had pulled the halliard so tight that the rope jumped the pulley up at the truck. Attempting to fix it this afternoon entailed removing the top sections of pole. Pulling on the weathered sash cord to free it only resulted in its breaking, which was probably the best result I could have hoped for.

So off I went to the hardware store and bought some new rope. It was only the work of a few minutes to cut the rope to size, attach the clips, and put the pole back up. Up went the flag as quick as I could raise it, then down slowly, very slowly, to half mast. And that's that.

You never know what kind of odd jobs you'll have to do in a given day, from explaining the doctrine of the holy Trinity to a bit of ropework. Ministry will take everything you've got that you can give.
moonbat alert

They must be Democrats

The problem given was, there are eleven members of a political party organization: 3 black males, 2 black females, 2 white males, and 4 white females. Party rules dictate that of the 3 officers, one must be black and one must be female. How many possible combinations of the members yield that result?

Well, I don't know the formula, but here's my work.

Call the 3 black men A, B, and C.
Call the 2 black women D and E.
Call the 2 white men F and G.
Call the 4 white women H, I, J, and K.

In order for a white male to be an officer, one of the other officers must be black. If the second officer is a black male, then the third officer must be a female; if the second officer is a black female, then the third officer can be any remaining member of the committee.

F or G + A requires D, E, H, I, J, or K = 12 combinations
F or G + B requires D, E, H, I, J, or K = 12 combinations
F or G + C requires D, E, H, I, J, or K = 12 combinations
F or G + D requires E, F or G, H, I, J, or K = 12 combinations
F or G + E requires F or G, H, I, J, or K = 10 combinations
Total combinations allowing for one or two white males = 58 combinations

Assuming that no white male is an officer, we start with one officer being a black male. The second officer must then be female. The third officer can be anyone except a white male (whose combinations are all used up in the first section).

A + D requires B, C, E, H, I, J, or K = 7 combinations
A + E requires B, C, H, I, J, or K = 6 combinations
A + H requires B, C, I, J, or K = 5 combinations
A + I requires B, C, J, or K = 4 combinations
A + J requires B, C, or K = 3 combinations
A + K requires B or C = 2 combinations
Total combinations allowing for black male A = 27 combinations
Substituting B for A will yield the same results, less 6 combinations already used = 21 combos
Substituting C or A will yield the same results, less 6 combinations already used = 21 combos
Total combinations allowing a mix of black males and white or black females = 69 combinations

Assuming an all-female officer cadre yields the following combinations.

D or E (1 black female) requires 2 white females in combos H and I, H and J, H and K, I and J, I and K, or J and K = 12 combinations
D + E (2 black females) requires 1 white female H, I, J, or K = 4 combinations
Total combinations for all-female officer cadres = 16 combinations

58 combos involving white males + 69 combos involving black males and white or black females + 16 combinations involving black females and white females = 143 possible combinations