combi

There are 2 possible cases,

Case1- Five 1's,one 2's,one 3's No.=7!÷5!=42

Case 2- Four 1's,three 2's No.of numbers=7!÷4!3!=35

Total no. Of numbers= 42+35=77

Done using generating functions!

${a}_{1} + {a}_{2} + {a}_{3} + ... + {a}_{7} = 10$ for ${a}_{i}$ s in [1,3]

${(x+{x}^{2} + {x}^{3})}^{7}$ is our function and we have to find the co-efficient of ${x}^{10}$

${x}^{7}({(1+x + {x}^{2})}^{7})$

So, we just have to find co-efficient of ${x}^{3}$ in $({(1+x + {x}^{2})}^{7})$

${(1-{x}^{3})}^{7}({(1-x)}^{-7})$

In the first, we will just get minimum of ${x}^{3}$ [except that one] and its co-efficient is $-C(7,1)$

In the second, we will get the coefficient of ${x}^{3}$ is $C(7 + 3-1,3) = C(9,3)$

Therefore, our answer becomes $C(9,3) - C(7,1) = 77$

My answer is same as that of Raghav Gupta , probably better presented here.

There are only $2$ possible cases:

• Case 1: $1+1+1+1+2+2+2=10$ . The number of seven-digit numbers of this case:

$\quad \quad N_1 = \dfrac {7!}{4!3!} = 35$

• Case 2: $1+1+1+1+1+2+3=10$ . The number of seven-digit numbers of this case:

$\quad \quad N_2 = \dfrac {7!}{5!} = 42$

Therefore, the total number of seven-digit numbers is:

$N= N_1+N_2 = 35+42 = \boxed{77}$