Question

Find the number of 7-digit numbers possible whose sum of digits is 61.
A. 12
B. 24
C. 28
D. none

Answer 1

Hint:- As we know that each digit can be from 0 to 9 but if all digits are nine then the sum will be 7*9 = 63 and if each digit if the number is eight then the sum of the digits will be 7*8 = 56. Use these two conditions to form the cases where the sum of the digits of 7 digit number is 61.

Complete step-by-step answer:
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 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, 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 \[{i^{th}}\] person is given \[{m_i}\] balls then the number of ways for this distribution is given as \[\dfrac{{n!}}{{\left( {{m_1}!} \right) \times \left( {{m_2}!} \right) \times ..... \times \left( {{m_r}!} \right)}}\]
And we know that if r is any positive integer than \[r! = r \times \left( {r - 1} \right) \times \left( {r - 2} \right) \times ..... \times 2 \times 1\]
So, therefore there are two possible cases.
Case 1:- ( five digits are 9 and two digits are 8)
So, number of possible numbers from this case will be \[\dfrac{{7!}}{{5!2!}} = \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{5 \times 4 \times 3 \times 2 \times 1 \times 2 \times 1}} = 21\]
Case 2:- (six digits are 9 and one digit is 7)
And number of possible numbers with this case will be \[\dfrac{{7!}}{{6!1!}} = \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 1}} = 7\]
Hence, the total possible even digits number with sum of its digit as 61 will be 21 + 7 = 28.
Hence, the correct option will be C.

Note:- Whenever we come up with this type of problem then we should check whether the sum of all digits is greater than the given sum if the value of each digit is greatest. From this we will get a minimum number of digits that should be equal to 9 to get the required number after that we fill the left digits such that their total sum is equal to the given value. And after that on applying formula for distribution of identical items (digits) on each case. We will get the required value of total numbers that are possible.