Let P be the set of seven-digit numbers with sum of digits equal to 11, formed using only the digits 1, 2 and 3. The number of elements in P is:
Solutions · 1
Official solution
We need 7 digits from {1,2,3} summing to 11. Since minimum sum (all 1's) is 7, we need 4 extra distributed as increments of 1 (to 2) or 2 (to 3).
(i) Digits: five 1's, one 2, one 3 i.e. 1,1,1,1,1,2,3 (sum =5+2+3=10, doesn't fit --- recompute): the valid multisets summing to 11 with 7 digits from {1,2,3}:
- {1,1,1,1,1,3,3}: sum =5+3+3=11 , arrangements =5!2!7!=21
- {1,1,2,2,2,2,3}: wait check sum =2+8+3, need recompute carefully --- using the source's valid cases:
- 1,1,1,1,1,3,3: 5!2!7!=21
- 1,1,1,2,2,3,… (two 2's, one 3, four 1's): 1,1,1,1,2,2,3: sum=4+4+3=11 , arrangements =4!2!7!=105
- 1,1,1,2,2,2,2: sum=3+8=11 , arrangements =3!4!7!=35
\[\text{Total} = 21+105+35 = 161\]