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Jessie and Casey are getting married. In how many ways can a photographer at their wedding arrange 6 people in a row from a group of 10 people, where the spouses are among these 10 people, if both spouses must be in the picture?

**My Attempt**

- discrete math>Recurrence relation>how find the general function of $a(n)=2a(n-1)+n^2$
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We are looking at $6$ seats, and $2$ of them must be filled by the spouses. The number of ways in which we can pick 2 seats out of $6$ is $P(6, 2)$ (I think a case can be made for both permutation and combination). For each permutation of the $2$ spouses, we have $4$ seats left and $8$ people to fill them so the final answer is $P(6, 2) \cdot P(8, 4)=50400$. However, this doesn’t correspond to any of the answer choices, which are

$a) 9 · 8 · 7 · 6 · 5$

$b) 8 · 7 · 6 · 5$

$c) 8 · 7 · 6 · 5 · 4$

$d) 8 · 7 · 6$

I tried the same method but with combinations and it’s wrong too. The choices would make sense if they specified that the spouses must sit next to each other in a specific order; in that case I think the answer would be $a)$, but they do not seem to make any such claims.

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Choose a place for Jessie ($6$ options) then for Casey ($5$ options) then fill the remaining places one by one – $8,7,6,5$ options in turn. The number of photo arrangements is thus $$8\cdot 7\cdot 6^2\cdot 5^2 = 50400$$

So none of the options given are correct as the problem is stated.

It’s sometimes usual to have the happy couple in a fixed position in the middle, in which case the initial choice of location for those two vanishes and we’re left with $$8\cdot 7\cdot 6\cdot 5 = 1680$$ possible arrangements around the couple.

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