Permutations Q (1 Viewer)

[flon]

In how many ways can a boat crew of 8 women be arranged if 3 women can only row on bow side and 2 others can only row on stroke side.

Thanks in advance

 

[Timothy.Siu]

4P3x4P2x3!

maybe

 

[hermine]

where did you get the 4's from ?

 

[Timothy.Siu]

where did you get the 4's from ?

4 positions for each side

 

[jpmeijer]

Assuming they have to be arranged 4 on bow, 4 on stroke,

no. of ways = (4.3.2)*(4.3)*3! = 1728

I did it by placing each person individually:

Firstly seat the 3 who must row on bow side:

1st one has 4 seats to choose from
2nd one has 3 seats to choose from
3rd one has 2 seats to choose from

Now seat the 2 who must row on stroke:

1st one has 4 seats to choose from
2nd one has 3 seats to choose from

Now there are 3 people left to be seated in the 3 remaining seats which can be done in 3! ways.

So just multiply all the bold numbers to get the answer.

Same as Tim's way really.

 

[flon]

Thanks. Helped a lot.

Heres another q thats bugging me.

In morse code letters are formed by a sequence of dashes and dots. How many different letters can be represented if a max of ten symbols are used?

The answer is 2046

 

[Timothy.Siu]

Thanks. Helped a lot.

Heres another q thats bugging me.

In morse code letters are formed by a sequence of dashes and dots. How many different letters can be represented if a max of ten symbols are used?

The answer is 2046

2^10+2^9+2^8+...+2=2(2^10-1)/(2-1)=2046

there might be a better way but it hasn't come to me yet...

basically there can be 2 symbols per slot, repeated

 

[jst.KP]

wow i jsut ran by this post and i had problems doing those exact two questions. . anyways..
can yu explain further the question on the morse code how yu did it/why

 

[exiting]

Lol I might confuse you, but my teacher told me instead of assigning people to seats, assign the seats to people. If I'm correct, this is from the year 12 Cambridge 3 unit book? Anyway, good luck!

 

[Prashant Sallan]

how do you know if you use permutations or combinations... it always confuses me..

 

[Timothy.Siu]

wow i jsut ran by this post and i had problems doing those exact two questions. . anyways..
can yu explain further the question on the morse code how yu did it/why

umm basically the question is asking how many things u can make if you are allowed to use a maximum of TEN symbols, this means u can use nine or eight or seven..etc.

and u have 2 "objects", either a dot or a dash to assign each "spot".

so, if u want to find the number if u use TEN symbols, any of the objects can go in each slot. so thats 2^10 different ways.
similarly for 9 symbols, its 2^9

so...if u want them from ONE symbol to TEN symbols,
2^10+2^9+2^8....+2=2046, using geometric sum

how do you know if you use permutations or combinations... it always confuses me..

if they ask for arrangments them its definitely perms, if it asks for just groups or something then its combs

 

[Drongoski]

Timothy.Siu

btw 2046 = 2¹¹ - 2 if that is clue to anything.

 

[jst.KP]

question on circular permutations

im very noob at circular perms =='

The letters A, E,I,P,Q,R are arranged in a circle.Find the probability that
i)A is opposite R
ii)at least two vowels are next to one another

 

[jst.KP]

btw thanks timothy ! i understand that now

 

[lychnobity]

question on circular permutations

im very noob at circular perms =='

The letters A, E,I,P,Q,R are arranged in a circle.Find the probability that
i)A is opposite R
ii)at least two vowels are next to one another

i) I'm not too sure on this one: (4! x 6)/5!

ii) Nor this one

Ways 2 vowels together = 3! x 3C1 x 2 = 36

Ways 3 vowels together = 2! X 3C1 x 3! = 18

P(at least two vowels are next to one another) = (36 + 18)/5! = 9/20

Somebody check if I'm right

 

[Timothy.Siu]

question on circular permutations

im very noob at circular perms =='

The letters A, E,I,P,Q,R are arranged in a circle.Find the probability that
i)A is opposite R
ii)at least two vowels are next to one another

at the guy above, ur answer in i) is wrong lol, ur answer gives 1.2 =P


i)hmm (6!/6)/5 x 2 all the possible arrangments divided by 5 (chance of R being opposite A) but x2 coz theres A and R?
i'm not sure though
or if u set the A and R, 4!x2? since A and R can swap places. lol

ii)well, 6!/6 - 3x3x2x2x1x1
i'm not sure either

 

[gurmies]

For the first part, is the answer 0.2 or 0.4? Not sure if I have to multiply by 2! in order to account for A and R swapping, because circle arrangement does some messed up changes to permutation theory -.- I'm sticking with 0.4 =D

 

[Timothy.Siu]

For the first part, is the answer 0.2 or 0.4? Not sure if I have to multiply by 2! in order to account for A and R swapping, because circle arrangement does some messed up changes to permutation theory -.- I'm sticking with 0.4 =D

oh it asks for probability
-2 marks for me