A few days ago
i need help with this math problem(POW) help!?
a car holds 6 people, in front seat and 3 in back seat. How many seating arrangements of the 6 people are possible if one person refuses to sit in the front and two other people refuse to sit in the back?assume al 6 are liscensed drivers.
Top 1 Answers
A few days ago
Favorite Answer
Oh, man. I am really no good at this, but here is my attempt:
People are:
1 – sits anywhere
2 – sits anywhere
3 – sits anywhere
4 – front only
5 – front only
6 – back only
Front seat has 18 possible combinations:
451 415 145
452 425 154
453 435 245
541 514 254
542 524 345
543 534 354
Back seat has 18 possible combinations:
612 126 216
613 162 261
621 136 236
623 163 263
631 316 361
632 326 362
Front and back seat combos are, I think,
18 times 18, or 324 possible seating arrangements.
I’m sure that there is some kind of formula to figure this kind of thing out, but I have no idea how to figure it!
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