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In how many ways can a photographer at a wedding arrange 6 people in a row from a group of 10 people, where the bride and the groom are among these 10 people, if

a) the bride must be in the picture?

b) both the bride and groom must be in the picture?

c) exactly one of the bride and groom is in the picture?

I'm having trouble intuitively understanding how this works. Can someone explain to me what it means when the "bride must be in the picture, etc." and how this effects the number of arrangements?

Thanks you.

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For a, you must have the bride and five others (one of whom may be the groom). You can select the five others in ${9 \choose 5}$ ways.Then arrange them in a row, which you can do in $6!$ ways.For b, you must have the bride, the groom, and four others, who you then arrange. Clearly there are fewer choices for the people than in a because one of the five must be the grrom. For c, you choose one of the bride or groom and five of the other eight.

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  • $\begingroup$ Hi, I asked a question about part $b$ of this question. Would you mind explaining it in a little more detail? math.stackexchange.com/questions/1828932/… $\endgroup$ – Ovi Jun 16 '16 at 19:46
  • $\begingroup$ @Ovi: it looks to me that Joffan did a good job. $\endgroup$ – Ross Millikan Jun 16 '16 at 21:20
  • $\begingroup$ Thanks for looking, I posted this before Joffan answered it $\endgroup$ – Ovi Jun 16 '16 at 23:26

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