QUESTION

# Find the number of seven-digit integers, with sum of the digits equal to 10 and are formed by using the digits 1, 2 and 3 only?  (A) 55  (B) 66  (C) 77  (D) 88

Hint- The question requires the concept of permutation and combination. In this question, first we will try to make different cases for seven-digit integers such that the sum of its digits is 10 and the digits are 1, 2 and 3 only. After that we will calculate the number of ways in which each case can be arranged.
Complete step-by-step solution -
Now, seven places need to be filled with 1, 2 or 3 and the sum of the digits needs to be 10.
_ _ _ _ _ _ _
Let us consider the case 1,
1 1 1 1 1 2 3
Here the sum of digits is equal to 10.
Now, consider the case 2,
2 2 2 1 1 1 1
Here the sum of digits is equal to 10.
Now, number of ways of arrangement of case 1 $= \dfrac{{7!}}{{5!}} = \dfrac{{7 \times 6 \times 5!}}{{5!}} = 7 \times 6 = 42$
Also, the number of ways of arrangement of case 2
$= \dfrac{{7!}}{{3!4!}} = \dfrac{{7 \times 6 \times 5 \times 4!}}{{3!4!}} = \dfrac{{7 \times 6 \times 5}}{{3!}} = \dfrac{{7 \times 6 \times 5}}{{3 \times 2}} = \dfrac{{7 \times 6 \times 5}}{6} = 7 \times 5 = 35$

Now, total number of seven-digit numbers formed = number of ways of arrangement of case 1 + number of ways of arrangement of case 2
$= 42 + 35 = 77$

Note- In these types of questions, always try to obtain different possible ways (cases) and their arrangement using permutation, half of the question will be solved by then. Also, No. of ways of permutations is denoted by ${}^n{P_r}$ where n represents the number of items to choose from, P stands for permutations and r stands for how many items we are choosing. Formula to calculate ${}^n{P_r}$ is ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$