Observations: consider wed+tab+pin=1**2

If we add up both alphametics we must get a number ending by 3
(13,23,33,43).

It is quite easy to understand that this number is either 23 or 33, since a sum
of 6 biggest digits is at most 39
-less than 43 , and the sum of six smallest, zero included is 15, more
than 13.

1. W+D+T+B+P+N=23 OR 33

2. 23 subdivides into 11+12
so both the 1^{st} sum and the second will begin by 1

3. So W+D+T+B+P+N= 33

4. The subdivision must be
D+ B+N=22 and W+ T+ P=11, leaving A+E+I =10 or 11

5. Trying 1hh2 , using the
same digit and trying leading zeros nulls immediately those options, - no need
to disclaim them a priori.

6. Same thing with VOWELS A+E+I=10 cannot produce a sum and its
reversal.

7. So be D+B+N=22 and W+ T+
P=11, leaving A+E+I = 11

8. Total of WEB PANDIT IS 44 i.e. DIGIT 1 NOT USED

It boils down to solving the basic set as follows:

wed+tab+pin=1232

dew+bat+nip=2321

W+D+T+B+P+N= 33, remember the subdivision

I=0; ARBITRARY CHOICE

A+E =11.

9. Trying 2+9, 3+8, 4+7, 5+6 for A+E we get generic solutions in no time –
presented in my previous post,-

Chose two generic solutions , and remember
that each represents 864 permuted versions.

<o:p> </o:p>

So the answer to a modified version should
refer to this number: **1728<o:p></o:p>**

I
honestly apologize for the condensed explanation- I am sure that whoever

tried to solve this puzzle gets a coherent
expose.

It is significantly easier to solve than to provide nicely structured
explanation.

I will sum up, how in my eyes this puzzle
should be rectified and

I
suggest you try to solve it from this text:

<begin <**d3**

**A set of alphametics**

WED + TAB + PIN = 1##2 or 2##1

DEW + BAT + NIP = 1##2 or 2##1,

in which "#" may represent any digit, has numerous
solutions.

Find , how many distinct solutions are there.

<end><o:p></o:p>

*Optional
Rem: No need to solve the alphametics*<o:p></o:p>

*Another
wording for puzzle's question" Using each of
the distinct solutions of the above set - in how many different ways can
you code " WEB PANDIT"? *<o:p></o:p>

<o:p></o:p>

Ans:
1728<o:p></o:p>

WEBPANDIT
:(SET1) &(SET2)- compare my prev post.<o:p></o:p>

Again
, excuse my shorthand style, using both capital and regular letters for the
same<o:p></o:p>

Item,
not editing for punctuation etc.<o:p></o:p>

<o:p> </o:p>

<o:p> **Errrare humanum est, perseverare diabolicum.**</o:p>

<o:p> </o:p>