Solutions
As we know that if all gifts are maximum (i.e., 9) then the sum of digits is 63 which is 2 more than 61.
So, all digits cannot be 9.
But if all digits are the second-largest digit (i.e., 8 ) then the sum is 56 which is 5 less than 61.
So, out of seven digits, five digits must be equal to 9.
So, the sum of five digits will be equal to 5×9=45.
So, the sum of the remaining two digits must be equal to 61−45=16.
Now as we know that if the sum of two digits is 16 then these two digits must be 8 and 8 or 9 and 7.
Now as we know that according to the identity of permutation which states that if there is are r persons and total n identical items and all n items are distributed to r persons and let each of the itℎ person is given m1 balls then the number of ways for this distribution is given as 

So, therefore there are two possible cases.
Case 1:- (five digits are 9 and two digits are 8)

Thus, the total possible even digits number with the sum of its digit as 61 will be 21+7=28.
Hence, the correct answer is 28