A few days ago
victor b

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
cardtapper

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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