CRYPTIC CHANCE

I'm a-gonna Fuddge on that one.

and nudge nudge wink wink

4 of the 6 squares can be selected in C(6,4)=15 ways.
Once the squares are chosen, they can be populated in
4! / (w! * x! * y! * z!) ways
where w, x, y, and z are the number of occurrences of the first, second, third, and fourth letter, and w+x+y+z=4
Obviously, if at least one letter is duplicated, there can't be four different letters - at least one of w, x, y, and z must be zero. But that's OK because 0!=1.

The answer you seek is the product of the two calculations. Assuming the four letters are unique, the answer is 15*24=360.

The general formula is easier to state when the letters are unique:
N = the number of squares
M = the number of (unique) letters

The number of ways is
C(N,M) * M! = N! / [(N-M)! * M!] * M! = N! / (N-M)!

twice the blanks were used
perfect cover for
darting letters

Huh?

the left hand column reads twice perfect darting and the answer is 360

sorry i forgot that your inclusion of the general theorum meant you get to eschew cryptic cruciverbalist stuff

congratulations

how about you resurrecting tryagain's page (that lad knew how to cut and paste) by posting an easy one so engineer and i, et al can battle it out?

Ah! A Ton-80 times 2. I'm a former darter, but I didn't catch the intended puzzle.

It appears that Tryagain is back. I don't think I could do justice to his page - he is one-of-a-kind.