IN A GAME OF POKER

There are three one eyed royals, two jacks and one king, so there are six full house combinations of interest:

1) 3K, 2X
2) 3K, 2J
3) 3J, 2 anything
4) 3X, 2K
5) 3X, 2J

Where X stands for a card not a jack or king. The possible combinations for each group are:

1) 4!/3! x 44x3/2! = 264 BUT ONLY THOSE WITH ONE EYE COUNT. For the kings, three out of four three card combos work, so 264x3/4 = 198.
2) 4!/3! x 4!/2!2! = 24 but once again, only those with one eye count. For the kings, that is 3/4 and for the jacks, that is 5/6, so 23 of the 24 hands work.
3) 4!/3! x 48x3/2! = 288. All hands with three jacks must contain a one eyed jack.
4) 44x3x2/3! x 4!/2!2! = 264, but only half will contain the one eyed king, so 132.
5) 44x3x2/3! x 4!/2!2! = 264, but only 5/6 will contain a one eyed jack so 220.

That totals up to 861 possible hands.

Thanks engineer, although the question was inadvertently more difficult than I had intended because I failed to constrain the results to full houses containing all the one- eyed royals, (cf. my pitiable didactics on other threads).

My solution to fullhouses with any one-eyed royal is:

With a king triple and any other pair there are

1C1 . 3C2 . 12C1 . 4C2 ways of including the 1 one-eyed king and any 2 of the other 3 kings together with 1 of the 12 possible pairs which require 2 of the 4 cards of that denomination ie

1 . [(3 x 2)/(1 x2)] . 12 . [(4 x 3)/(1 x 2)] = 1 x 3 x 12 x 6 = 216

With a jack triple and any other pair there are

[2C2 . 2C1 + 2C1 . 2C2] . 12C1 . 4C2 ways of including [ either 2 one-eyed jacks and another jack or 1 one-eyed jack and the other 2 jacks] together with 1 of the 12 possible pairs which require two of the four cards of that denomination ie

{[(2 x 1)/ (1 x2)]. 2 + [ 2.(2 x 1)/(1 x 2)]}. 12 . [(4 x 3)/(1 x 2)] = 4 x 12 x 6 = 288

ie 504 combinations

With a king triple and any other pair there are

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coincidence and kismet are like odds and probability