This has 864 permutation solutions for which there are 2 permuted solutions.
When written as:
W E D and D E W
T A B B A T
P I N N I P
X j k Y Y k j X
it quickly becomes obvious that X can only be 1 or 2, and so by reflection then Y is whatever X is not.
If X = 2 then W + T + P = 22 to take into account that upon reflection the thousands and units digits must be 2.
Similarly if Y = 1 then D + B + N = 11.
For a total of 22 each of the digit addends must be at least 3 which precludes X and Y in this case.
22 will be the sum of the digits of either 9,8 & 5 or 9,7 & 6.
The sum E + A + I cannot equal 1, and once the sets of 22 and 11 are satisfied these remaining digits must total 11.
Compiling these place value totals, 22,11,11 and 11,11,22
renders:
X j k Y = 2321 or 1232
and Y k j X = 1232 or 2321
If 11 has 1 within is summation it is formed from the digits: 1,2,8; 1,3,7 or 1,4,6.
Paired with the make-up of the 22 totals, and the remaining digits(R), the following table offers:
(11) (R) (22) (11) (R) (22) (11) (R) (22)
1 0 9 1 0 9 1 0 9
6 2 8 7 2 8 8 3 7
4 3 5 3 4 5 2 4 6
7 6 5
In all cases remaining digits(R) total 12!
Dismissing the "1s" in the (11) columns and swapping the underlined digits (1 for 2) in each case forms the balance 11,11,22.
1 cannot appear as any of the digital addends for 11.
While it may be noted that 2 can occur as an addend for 11 the problem did state:
"Why can X and Y never equate to ... ".
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