## Saturday, 23 April 2011

### References + more studies....

Well, I'm a month away from the Fulbright scholarship deadline and have just been pinged a note by the system that two out of three references are secured + submitted! :)  Next week I'll polish off the remaining essays I've previously worked up to a decent draft stage, leaving just the final reference to chase.  No sweat!

These last few days I've been hammering the quant questions available through the forums on http://www.gmatclub.com/ - lots of fun (!).  Was stumped by one particular combination question that asked:

"There are three secretaries who work for four departments. If each of the four departments has one report to be typed out, and the reports are randomly assigned to a secretary, what is the probability that all three secretaries are assigned at least one report? "

What had me really stumped was that once I thought I had the answer, it was significantly different from the OA.  As the OA looked a bit dodgy, I thought I'd trawl google for suggested solutions and answers and was somewhat dismayed to find a number of different answers posted by various tutors and quant-proficient individuals.....

Being something of an excel geek, I then decided to check this out long-hand myself.... as I thought, the OA offered in the gmatclub download was wrong.  Nevermind, the combinatronics document is otherwise a good resource and my the learning experience has been enriched somewhat by having to prove why my thinking was right :) :) here's my final answer ......

Spoiler:  There are 3^4 (or 3*3*3*3) possible combinations = each option therefore is 1/81.

You then choose 1 of three secretaries (3C1) to receive two reports (4C2), and then work out the number of permutations to assign the remaining 2 reports to the 2 remaining secretaries (2P2).

3C1 * 4C2 * 2P2 = 36

Now it's time to cycle down to the lake, enjoy the sunshine and run through some DS problems.