That's the only numbers I know

Combinations are 1 111 123---a, and 1 111 222---b only. So number of permutations are a: 7!/5!, and b: 7!/3!4!=42+35=77

co-efficient of x^10 in the expansion of (x+x^2+x^3)^7

First, we make the choice of numbers. The only possible choices are $1, 1, 1, 1, 1, 2, 3$ and $1, 1, 1, 1, 2, 2, 2$ . Why these are the only two possible choices is a little hard to explain, but think about it logically and you will understand.

The number of numbers possible in the first case is $\frac{7!}{5!} = 42$ . Here, $7!$ represents the number of ways to arrange 7 things and we divide it by $5!$ because the 5 "ones" arranged in any way are the same. Similarly, the number of numbers possible in the second case are $\frac{7!}{3! \times 4!} = 35$ . Here, we are dividing by both $4!$ and $3!$ to account for the "ones" and the "twos".

So, the total no. of numbers possible is $42 + 35 = \boxed{77}$

That is the same as having 7 urns and 10 balls, and you want to put at least 1 ball, but no more than 3, in each urn.

As there's at least one ball, you put one in each urn and you're left with 3 balls to put wherever you want, but not all in the same urn.

There's $\Big(\!\!\Big(\begin{matrix}7\\3\end{matrix}\Big)\!\!\Big)={7+3-1\choose3}=84$ ways to put the 3 balls, but we don't want the 7 ways where all are put in the same urn. That leaves $84-7=\boxed{77}$ .

$\Big(\!\!\Big(\begin{matrix}n\\p\end{matrix}\Big)\!\!\Big)$ is the multicombination: choosing $p$ times from $n$ elements without order where it is allowed to choose the same element multiple times.

As the sum of digits has to be 10 and each digit is out of {1, 2, 3}.

Let's think this in a different way. There are 7 children and we have 10 chocolates, we need to distribute all the chocolates to all the children. In how many ways we can do that ?

As each child must get at least 1 chocolate, we will distribute each child 1 chocolate. Now each 7 children has 1 chocolate in their hand. Now we have remaining 3 chocolates in our hand and we would like to distribute to the children.

• Give 3 chocolates to 3 different child. That can be done in ${7 \choose 3}$ ways.
• Give 2 chocolates to one child and 1 to another child. That can be done in $^{7}P_2$ ways.

$\therefore$ The answer is ${7 \choose 3} + {^{7}P_2}$ = 77.

The only 2 possible formats:

a) 5 1's, 1 2, and 1 3, e.g., 1111123, or 1121131

b) 4 1's, and 3 2's, e.g., 1111222, or 1212121

For a), there are $7 \choose 2$ * 2! = 21 * 2 = 42 ways.

For b), there are $7 \choose 3$ = 35 ways.

In total, there are 42 + 35 = 77 ways

Python:

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from itertools import product
def digit_sum(string):
    total = 0
    for digit in string:
        total += int(digit)
    return total
count = 0
for combo in product('123', repeat=7):
    test = ''.join(combo)
    if digit_sum(test) == 10:
        count += 1
print "Answer:", count