Thank you, Omi67!
The total number of ways of seating \(7\) people is \(7!\) The only disallowed seating arrangement is alternate boy-girl, \(\texttt{BGBGBGB}.\) There are \(4!\) ways to seat the boys and \(3!\) ways to seat the girls. So the number of disallowed seating arrangements is \(4! \times 3!\) The final answer is \(7! - 4! \times 3! = 5040-24\times 6=5040-144=4896\)
Happy belated birthday, Melody! A great asset to the forum, and hope you continue to grind!
Thanks for the detailed explanation, Gavin!
https://web2.0calc.com/questions/on-the-student-government-ballot-at-dusable-high-school-the-six-candidates-for-president-are-listed-first-followed-by-the-four-candidates-for-vice
6!*4!*5!*3!
We break \(27^x\) into \((3^3)^x\) , and \(9^4\) into \((3^2)^4\) . Now, we have \(3^2*3^8=3^{3x}\) . Since the bases are equal, we have \(3x=2+8, 3x=10, x=\boxed{\frac{10}{3}}\)
Multiples of 3: 100/3= 33 1/3 - 33
Multiples of 4- 25, some are multiples of 3, so 25-8=17, so 33+17=50
Wait, so the total is 60, right? Okay \( x^2-2x-x+2x+8-x+x+10=60\)
Same here! I'll try my best!
Hmm, I think you're correct: \(\frac{13}{35}+\frac{2}{5}=\boxed{\frac{27}{35}}\)
