How many non-negative integer solutions are there for the following?
$$a + b + c = 5$$
From my understanding, the answer would be:
$$\frac {7!}{5!\cdot 2!} = 21$$
I tried applying the same principle to the question below.
There are 5 married couples. How many ways can we arrange them in a line if none of the husbands are next to each other?
Method 1
Let $H$ represent a husband and $W$ represent a wife.
Thus, the 5 couples can be arranged in the following manner:
$$\text {_} \; W_1 \; \text {_} \; W_2 \; \text {_} \; W_3 \; \text {_} \; W_4 \; \text {_} \; W_5 \; \text {_}$$
The _ represents a position where a husband can be placed.
The wives can be arranged in $5!$ ways.
There are 6 positions that $H_1$ could take, 5 positions that $H_2$ could take, 4 positions that $H_3$ could take, and so on, which leads to $6\cdot 5\cdot 4\cdot 3\cdot 2 = 720$ total ways to arrange the husbands.
Thus, the answer is $5!\cdot 720 = 86400$ ways to arrange the 5 couples.
Method 2
Looking back at the following:
$$\text {_} \; W_1 \; \text {_} \; W_2 \; \text {_} \; W_3 \; \text {_} \; W_4 \; \text {_} \; W_5 \; \text {_}$$
If each _, of which there are 6 in total, represents a variable such that $a + b + c + d + e + f = 5$, there would be $\frac {10!}{5!}$ ways to arrange the 5 husbands.
This leads to a total of $\frac {10!}{5!} \cdot 5! = 10! = 3628800$ ways to arrange the 5 couples.
Why is the second method incorrect?