Mathematics: Post your doubts here!

[Khan_971]

and please i need help in question 6 part iv) and v) in nov 10 paper 61 ??
thnx

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.

 

[Khan_971]

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

 

[user]

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

 

[Khan_971]

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

 

[omar hazem]

usually.

ok when can i know exaclty to use the cc, this may lead to losing free easy marks :/

 

[Khan_971]

ok when can i know exaclty to use the cc, this may lead to losing free easy marks :/

the only situation ive encountered it in is when 'at least' or 'not more than'

 

[Wanzi21]

what to do on the 2nd part?

 

[omar hazem]

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??

 

[smartangel]

) 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!

 

[Khan_971]

View attachment 10573

what to do on the 2nd part?

Add (29-45) to -148
add (29-48)^2 to 3089.
and carry on

 

[Khan_971]

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 :/

 

[alphabravocharlie]

a.
52^8 = 5.346x10^13

b.
12^3/3! * 20^3/3! * 20^2/2! * 8! = 3.097x10^12

uhm, could you explain part b please

 

[Wanzi21]

Add (29-45) to -148
add (29-48)^2 to 3089.
and carry on

you are really a master!!
thanks

 

[asexamskillme111]

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?

 

[Khan_971]

uhm, could you explain part b please

Since there's replacement so ill not consider P or C. just poweres. since 12 pic cards are there and i have to pic 3. 12^3. dividing by 3! is to cut down the number of arrangements OF THIS SELECTION. same goes for the rest. 8! is an overall arrangement ( 8 cards)

 

[omar hazem]

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.

i think the last one we need to find 2 different colours which will be denoted by 4!/2!2! and as we need 2 for each colour so the 6 different colours is to be selected for 2 holes so 6c2 so the answer is 90 and then add all of the previous answers to it

 

[Khan_971]

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?

Explain this please (7*6*5*(6!/3!)).

 

[alphabravocharlie]

Since there's replacement so ill not consider P or C. just poweres. since 12 pic cards are there and i have to pic 3. 12^3. dividing by 3! is to cut down the number of arrangements OF THIS SELECTION. same goes for the rest. 8! is an overall arrangement ( 8 cards)

Oh! JazakAllah brother

 

[asexamskillme111]

Explain this please (7*6*5*(6!/3!)).

GGG REEENA. I did 7*6*5 as in 7 places for the first G, 6 for the second and 5 for the 3rd G. and 6!/3! is the number of arrangements of REEENA.

 

[alphabravocharlie]

GGG REEENA. I did 7*6*5 as in 7 places for the first G, 6 for the second and 5 for the 3rd G. and 6!/3! is the number of arrangements of REEENA.

What is the answer?