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How would you have done it?

wait now im confused. why isnt this correct? : 1. there are four different groups in the first row: 4! 2. there are also four different groups in the second row: 4! 3. the remaining 5 people can rearrange themselves in anyway: 5! 4!x4!x5!

oh i get it now! thanks guys :)

woops. yep, that's what i meant. which was why i didnt do 4C2

the answer says 5760 in terry lee

Well, i said Just pick any two boys, make them one group in the first row. So now there are 3 groups at the front. we can arrange the boys, 2! and the three groups, so 3! In the last row, there are 4 groups, so 4! . Besides the two boys at the front and the one girl at the back, everyone else...

waaaa ): bump

Four girls and four boys are to sit in two horizontal rows of four. find the number of permutations if A) Two boys want to sit in the front row, and a girl wants to sit in the back row Attempt: 2! x 3! x 4! x 5! So apparently my answer is wrong [5760]. Can someone please correct me?

): i think im just burning out and i need to relax. not the best timing right now T___T

i feel so unmotivated to study for my half yearlies.............

find all real x such that: |4x-1|>2sqrt[x(1-x)]

bump

determine for what value of c the equation |2x-1|=x+c has exactly two solutions