I'm trying to determine the game 5 card poker when order is matter all my trails ended with fail except the one pair. The way I did one pair assuming it is $\{k,k,x,y,z\}$
$^{13}P_1 \times ^4P_2 \times ^{12}P_3 \times (^4P_3)^3$
Taking 1 permutation out of 13 cards and then multiply it with how many suit permutation for the first 2. in this case we will have the possible hands of the first two cards with order. Then by multiply it with the second part of taking 3 different cards from the rest (12 cards) including the 4 different suits I got the right number. 13,178,880
Now by applying the same rule to Two-pair $\{kk,dd,x\}$ using two different ways.
1- $^{13}P_1 \times ^4P_2 \times ^{12}P_1 \times ^4P_2 \times ^{11}P_1 \times ^4P_1 = 988,416$
2- $^{13}P_2 \times(^4P_2)^2 \times ^{11}P_1 \times ^4P_1 = 988,416$
the result is wrong. Considering the permutation between the three group which means multiply the result by $^3P_3$ = 6 is equal to 5,930,496 is still wrong the correct result based on the book is 14,826,240.
My trials stopped just here waiting to find a way and the two find the answer for the rest of the game.
The calculation that got you 988416 is correct. However, there should be $\displaystyle \frac{5!}{2! \cdot 2! \cdot 2!}$ possible permutations for $\{kk,dd,x\}$.
$5!$ is the number of permutations for 5 cards, and the three $2!$'s in the denominator represents the numbers of permutations of the 2 $k$'s, the 2 $d$'s, as well as $kk$ and $dd$.
$$ 988416 \times \frac{5!}{2! \cdot 2! \cdot 2!} = 14826240 $$