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Mathematics: Post your doubts here!

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Find the number of different ways in which the 9 letters of the word GREENGAGE can be
arranged if exactly two of the Gs are next to each other.
Now in this question I understand the way to do it but when I was trying to do it in different way, I hit a wall. I know I'm doing something wrong I just can't figure what is it.
This is what I want to do:
Total number of arrangements - number of arrangements where none of the G's are next to each other.
9!/(3!*3!) - (7*6*5*(6!/3!)). THE ANSWER IS NEGATIVE. WHAT AM I DOING WRONG?
 
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i don't understand why sometimes we square or cube our answers? especially in those questions where they asked the probability of something happened for the next three days or next five churches? can someone please explain it for me? thanks
 
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and please i need help in question 6 part iv) and v) in nov 10 paper 61 ??
thnx :D
Part 4...hmmm. okay see there will normally be 4! arrangements if 4 diff. colors are used. since 2 are the same, we make it 4!/2!.
there are 6 colors in total, AND since u have to choose 1 peg of a specific colour only, 6x 5C2
therefore,
4!/2! arrangements * 6 colors * 5C2 pegs of diff colours = 720

part v)
im really not sure.
 
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i don't understand why sometimes we square or cube our answers? especially in those questions where they asked the probability of something happened for the next three days or next five churches? can someone please explain it for me? thanks
probability of sth to happen = x
when it is happening more than 1nce and being independent (no replacement/removal), then to happen next time, probability will be same
so if they tell u to find that that this happens twice. Probability will be x *x = x^2. if thrice, then x^3
 
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Aoa wr wb!

need help with part b

8 cards are selected with replacement from a standard pack of 52 playing cards, with 12 picture, 20 odd and 20 even cards.
a) how man different arrangememts of 8 cards are possible.
b) how many arrangements in a contain 3 picture, 3 odd and 2 even cards

Answer for first: 5.346 x 10^13
For second: 3.097 x 10^12
 
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Aoa wr wb!

need help with part b

8 cards are selected with replacement from a standard pack of 52 playing cards, with 12 picture, 20 odd and 20 even cards.
a) how man different arrangememts of 8 cards are possible.
b) how many arrangements in a contain 3 picture, 3 odd and 2 even cards

Answer for first: 5.346 x 10^13
For second: 3.097 x 10^12


a.
52^8 = 5.346x10^13

b.
12^3/3! * 20^3/3! * 20^2/2! * 8! = 3.097x10^12
 
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the only situation ive encountered it in is when 'at least' or 'not more than'
they said before in this thread if they said use suitable approximation or random but i understand from ur words that when the random variable X is normally distributed and it is at least or should not be greater we use the cc even if it is normal distribution, right??
 
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) Find how many different odd numbers greater than 500 can be made using some or all of the digits 1, 3, 5 and 6 with no digit being repeated.
ANS 28.. plz explain n if someone has then plz give me the link to that post!
 
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they said before in this thread if they said use suitable approximation or random but i understand from ur words that when the random variable X is normally distributed and it is at least or should not be greater we use the cc even if it is normal distribution, right??
CC is only in normal approximation :/
 
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Find the number of different ways in which the 9 letters of the word GREENGAGE can be
arranged if exactly two of the Gs are next to each other.
Now in this question I understand the way to do it but when I was trying to do it in different way, I hit a wall. I know I'm doing something wrong I just can't figure what is it.
This is what I want to do:
Total number of arrangements - number of arrangements where none of the G's are next to each other.
9!/(3!*3!) - (7*6*5*(6!/3!)). THE ANSWER IS NEGATIVE. WHAT AM I DOING WRONG?

can someone reply please?
 
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