Q2
$${\frac{{\mathtt{9}}{!}}{\left({\mathtt{2}}{!}{\mathtt{\,\times\,}}{\mathtt{3}}{!}\right)}} = {\mathtt{30\,240}}$$ permutations.
------------------------------------------------------
Q5
KING Big square photo of himself and Smaller square on of wife Totally = 1865cm^2
QUEEN Big square phot of herself and tiny square on of husband Totally = 1865cm^2
The lengths of all the photos is an exact number of cm.
$$x^2+y^2=1865$$
We can solve this with just a little sensible trial and error.
$${\sqrt{{\mathtt{1\,865}}}} = {\mathtt{43.185\: \!645\: \!763\: \!378\: \!368\: \!2}}$$
So for the 2 that are a long way apart one is are going to be just a little smaller than this number and the other will be little.
Try 43
$${\sqrt{{\mathtt{1\,865}}{\mathtt{\,-\,}}{{\mathtt{43}}}^{{\mathtt{2}}}}} = {\mathtt{4}}$$ ok so the first 2 queens ones are 4cm and 43cm
Now lets think about the 2 that are close together.
think about
$${\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{x}}}^{{\mathtt{2}}} = {\mathtt{1\,865}} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = {\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{1\,865}}}}}{{\sqrt{{\mathtt{2}}}}}}\\
{\mathtt{x}} = {\frac{{\sqrt{{\mathtt{1\,865}}}}}{{\sqrt{{\mathtt{2}}}}}}\\
\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{x}} = -{\mathtt{30.536\: \!862\: \!969\: \!204\: \!940\: \!9}}\\
{\mathtt{x}} = {\mathtt{30.536\: \!862\: \!969\: \!204\: \!940\: \!9}}\\
\end{array} \right\}$$
okay so the two close together will be near 30
Try 30
$${\sqrt{{\mathtt{1\,865}}{\mathtt{\,-\,}}{{\mathtt{30}}}^{{\mathtt{2}}}}} = {\mathtt{31.064\: \!449\: \!134\: \!018\: \!134\: \!1}}$$
Try 29
$${\sqrt{{\mathtt{1\,865}}{\mathtt{\,-\,}}{{\mathtt{29}}}^{{\mathtt{2}}}}} = {\mathtt{32}}$$
Bingo, the Kings ones are 29cm and 32 cm
The King's pictures were 32cm x 32cm and 29cm x 29cm
and
The queen's were 43cm x 43cm and 4 cm x 4cm
Ik Q2
I'll do a simple example to show you
A1,A2,A3,B
how many ways can these be placed in a line 4*3*2*1=4! ways.
Now lets forget about the B for a moment, how many ways can the A's be ordered. that would be 3!=6 ways
So if you treat the As like they are all the same there will be 4!/3! ways of arranging them. That's 4 ways.
BAAA, ABAA, AABA, AAAB THAT IS IT.
-------------------------
now marmalade is made of m m a a a r l d and e
9 letters so if they were all different there would be 9! ways.
But
there are 2 Ms and 3 A's
so that is 9!/(2!*3!)
Hi MathsGod,
No,
9! means 9*8*7*6*5*4*3*2*1
Say you had the letters A,B,C,D,E,F,G,H,I
There are 9 letters.
So there are 9 letters to chose from for the first letter.
then there are 8 letters to choose from next
then 7 and so on untill there is just 1 then 0 letters left
so the number of ways that they can be ordered is
9*8*7.....*1 as I wrote above.
9!/(2!3!) = $$\frac{9!}{2!3!}=\frac{1*2*3*4*5*6*7*8*9}{1*2\;\;*\;\;1*2*3}$$ $${\frac{{\mathtt{9}}{!}}{\left({\mathtt{2}}{!}{\mathtt{\,\times\,}}{\mathtt{3}}{!}\right)}} = {\mathtt{30\,240}}$$
that page of questions that you found are far to hard for you, that is why you cannot understand my explanations.
If you would like to extend yourself that is excellent but you need to find something just a little different from what you are doing in class. Not something that is years above you. :)
How about if I find something for you.
Maybe you will enjoy this video clip.
https://www.khanacademy.org/math/precalculus/prob_comb/basic_prob_precalc/v/basic-probability
So altogether there would be 15?
2: 1*2 (2 options)
3: 1*2*3 (3 options)
9: 9*8*7*6*5*4*3*2*1 (9 options)
9+3+2=15? Sorry I'm not familiar with this working but I hope I'm getting it.
Also, thanks for the support it's much appreciated.
:)
But then 15 seems so small.
Is it 15 per letter. So 15*9=135
Sorry for all the Q.'s :(
Hi MathsGod
I do not mind you asking questions but i do not understand what your question is.
A * sign means multiply. :)
since it's 9!/(2!*3!) if i was to simplify it into one number. I thought it'll be 15 ? But it seems to small so for each letter? 15*9=135
no
$${\frac{{\mathtt{9}}{!}}{\left({\mathtt{2}}{!}{\mathtt{\,\times\,}}{\mathtt{3}}{!}\right)}} = {\mathtt{30\,240}}$$
Why don't you try a little one.
Say
AABBB
how many permutations can you make with those letters? Just copy my technique.
Can you name them all?
Yes great.
5! Is 5 factorial and it means
1x2x3x4x5
so what would 5!/(2!×3!) equal?
5!/(2!×3!)
1*2*3*4*5=120
1*2=2
1*2*3=6
is it 120 divided by 6 * 2 ? $${\frac{{\mathtt{120}}}{\left({\mathtt{6}}{\mathtt{\,\times\,}}{\mathtt{2}}\right)}} = {\mathtt{10}}$$
Excellent
now see if you can write out all those permutations.
there are exactly 10 of them 😉
Since theirs different types of A does that mean AABBB and AABBB could be different.
AABBB
ABBBA
BBBAA
BABAB
BAABB
BABBA
ABABB
ABBAB
BBABA
um... >.<
You are missing one MathsGod - you have to be very methodical
I am going to move the A's about in a very consistent way and hopefully I will get all 10 of them.
AABBB A is first
ABABB
ABBAB
ABBBA
BAABB B is first and A is second
BABAB
BABBA
BBAAB B is 1st and second and A is 3rd
BBABA
BBBAA B is 1st 2nd and 3'rd A is 4th.
And that is 10
Here is a new one if you would like to try it.
The word is "Look"
How many permuations of the letters in this word are there?
Can you list them?
I am MathsGod P.S
Look
ookl
4!/(2!)
1*2*3*4=24
1*2=2
24/2=12
Look
Lkoo
Loko
Okol
Olok
Oolk
Oklo
Ookl
Olko
Kool
Kloo
Kolo
Thanks.
Its easier when you start with one letter to focus on!
Is this right?
Yes, that 'looks' really good MathsGod
Here is a different question.
Dora has 2 brothers and 2 sisters. How many Permutation of the family can there be (from Dora 's point of view)
For instance Dora could have 2 older brothers and 2 younger sisters. That is one possibility. :)
When you get sick of answering these questions you can either tell me or you can just stop answering.
I won't mind :)
I don't mind answering this last one
2 brothers , 2 sisters ... is it BBSS?
Or since they could be either older or younger OB,YB,OS,YS
This is harder
BBSS=
$${\frac{{\mathtt{4}}{!}}{\left({\mathtt{2}}{!}{\mathtt{\,\times\,}}{\mathtt{2}}{!}\right)}} = {\mathtt{6}}$$
?
Older=O , Younger=Y
OB,OB,OS,OS
OB,OB,YS,YS
OB,OB,OS,YS
OB,YB,OS,YS
OB,YB,OS,OS
OB,YB,YS,YS
?
Ha,just realised i had only done the OB section
OB,OB,OS,OS
OB,OB,YS,YS
OB,OB,OS,YS
OB,YB,OS,YS
OB,YB,OS,OS
OB,YB,YS,YS
YB,YB,YS,YS
YB,OB,OS,OS
YB,OB,yS,YS
YB,OB,OS,YS
YB,YB,oS,YS
YB,YB,OS,OS
I think that's all.
OB,YB,OS,YS
$${\frac{{\mathtt{4}}{!}}{\left({\mathtt{1}}{!}{\mathtt{\,\times\,}}{\mathtt{1}}{!}{\mathtt{\,\times\,}}{\mathtt{1}}{!}{\mathtt{\,\times\,}}{\mathtt{1}}{!}\right)}} = {\mathtt{24}}$$
Or
BBSS
$${\frac{{\mathtt{4}}{!}}{\left({\mathtt{2}}{!}{\mathtt{\,\times\,}}{\mathtt{2}}{!}\right)}} = {\mathtt{6}}$$
But i think i got all the combinations right.?
Hi MathsGod,
You have stumbled on a little problem. I put it there to confuse you :))
I think you have missed out on some ot the answers.
We have B B S S and Dora
so I think that the number of permutations is $$\frac{5!}{2!2!}=30$$
Can you find them all?
Notice that I keep using the word PERMUTATIONS. This is a very specific maths word.
When doing probablilty it is very important to use the correct language, otherwise people will often misinterprete your meaning. :)
And Dora ...
How did I not notice?
Ha!
So,BBSSD
$${\frac{{\mathtt{5}}{!}}{\left({\mathtt{2}}{!}{\mathtt{\,\times\,}}{\mathtt{2}}{!}\right)}} = {\mathtt{30}}$$
Hmm...Is there a quick way to find the 'Permutations'
Have you had enough yet or so you want another? (I will let you finish this one first anyway)
I will do another one.
I want to try it without help :D
Also, is their best way to find the permutations just by doing it each letter at a time, carefully?
There is not really a quick way exactly - you need to be methodical.
I would make a list of the 4 without Dora first (olders to youngest) That is only 4!/(2!2!) = 6
then I woud repeatedly cut and paste that list putting Dora in each posible position.
so for the fist cut an paste Dora can be eldest.
for the next cut and past Dora can be second and so one.
So there will be 5 cut and pastes
5*6=30
so that would be pretty quick. Do you understand what I mean?
Well kind of.
4!/(2!*2!)=6
1.BBSS
2.SSBB
3.SBSB
4.BSBS
5.SBBS
6.BSSB
DBBSS
BDBSS
BBDSS 1
BBSDS
BBSSD
DSSBB
SDSBB
SSDBB 2
SSBDB
SSBBD
DSBSB
SDBSB
SBDSB 3
SBSDB
SBSBD
DBSBS
BDSBS
BSDBS 4
BSBDS
BSBSD
DSBBS
SDBBS
SBDBS 5
SBBDS
SBBSD
DBSSB
BDSSB
BSDSB 6
BSSDB
BSSBD
Is that all ?
I hope so >.<
Well 5*6 is 30 and that is what you were expecting so YES you got them all ant it was not really that hard either was it?
Here is one that is just a little different.
Suzie has 10 different coloured beads. She uses these beads to make a bracelet.
This bracelet does not have a clasp so there is really no beginning or end (but there is clockwise and anticlockwise)
How many different permutations of beads are possible ?
Try question 5 :)
http://www.regentsprep.org/regents/math/algebra/APR2/PracPerm.htm
I'm going to have to go after this question.
:)
digit=
iidgt
$${\frac{{\mathtt{5}}{!}}{{\mathtt{2}}{!}}} = {\mathtt{60}}$$
Wow, i am shocked how much with permutations i have improved on because of you Melody.
I wrote another question for you a few posts back.
I whited it our because it is a little different from the others.
Maybe you would like to look at it. I shall unwhite it for you.
It is the one about the bracelet. :)
Sorry i guess i didn't refresh the page ii had only seen the link.
So 10 different coloured bead . A,B,C,D,E,F,G,H,I,J. (Just to make it easier :) )
Is it just... Since non of the letters are the same.
$${\mathtt{10}}{!} = {\mathtt{3\,628\,800}}$$
and if i was do 1!,10x it would be same
3628800.Too much >.<
This was the question:
Here is one that is just a little different.
Suzie has 10 different coloured beads. She uses these beads to make a bracelet.
This bracelet does not have a clasp so there is really no beginning or end (but there is clockwise and anticlockwise)
How many different permutations of beads are possible ?
----------------------------------------
Your answer was 10!
If these beadswere placed in a row your answer would have been correct, but when you put the beads in a circle there is no distict start point so you have to put the first bead down to indicate first place. Now there are 9 beads (not 10 ) that need to be organised around this initial bead.
so the answer is 9! $${\mathtt{9}}{!} = {\mathtt{362\,880}}$$ so if you go clockwise (or anticlockwise) from any specific point there will be 362 880 possible permutations.
Hi Melody,
I'm having some trouble trouble with angles with my 'basic' level work at school.
I'm wondering how I'd figure out these angles on my online Homework.
I took some screen shots of it.
http://imgur.com/i4UbIPQ
PS. Don't worry about cheating each time the angle changes
If you could help me it would be much appreciated.
Thanks,
Question 2
Here are the answers for all of these
For the 1st one (with the 126 degree angle), x = 180 - 126 = 54 degrees (consecutive interior angles)
For the one with the 113 degree angle, x = 113 degrees (corresponding angles)
For the one with the 122 degree angle, x = 122 degrees (alternate interior angles)
For the one with the 80 degree angle, x = 180 - 80 = 100 degrees (x is supplemental to the vertical angle formed by the 80 degree angle.....thus this angle is a consecutive interior angle to x)
For the one with the 136 degree angle, x = 180 - 136 = 44 degree (x is supplemental to the vertical angle formed by the 136 degree angle.....like the last one, this angle is a consecutive interior angle to x)
For the one with the 61 degree angle, x = 180 - 61 = 119 degrees (same side exterior angles are supplemental)
Hi again MG1,
I only just saw you question and CPhill has already answered you. :)
I have found that CPhill useds slightly different language for these than me. It is a country based difference. He is from the US and I am from Australia. So his terms are probably better for you but I will run through it anyway.
1st on top row)
Cointerior angles (C shaped) These are supplementary angles which means that they add to 180 degrees
126+x=180
x=180-126
x= 54 degrees
2nd on top row )
Corresponding angles (F shaped) If you think about it you will see that they are corresponding because they are in the same position.
Corresponding angles on parrallel lines are congruent (the same)
x=113 degrees
last on top row.
Alternate (z shaped) if you think about it they zig zag. Alternate angles are congruent.
x=122
Second row.
I don't have specific names for these I would work them out with 2 of the above ones each.
The first one I would use corresponding then adjacent supplementary.
The answer would be 180-80=100 degrees
the second one I'd use the same, the answer will be 180-136= 44degrees
The third one id do the same, the answer will be 180-61=119 degrees
Well I understand better now but the questions that iim really struggling at are the ones involving Algebra.
http://imgur.com/i4UbIPQ
y=64 corresponding angles on parallel lines (F shape)
The one opposite y in the quadrilateral is 75 degrees because it is corresponding on parallel lines to the other 75.
x+75=180 because they are adjacent and supplementary
x=105 Adjacent because they share a common vertex and a common arm.
Adjacent is a normal English word, it means next to, so the property that is adjacent to yours is the one that shares the same fence line.
Supplemenary just means that they add to 180. (because they form a strght line)
----------------
the one opposite the x in the quadrilateral is 111 because it is a corresponding angle to 111 on parallel lines.
The one in the triangle next to it is 180-111=69 because they are adjacent supplementary angles.
the other one in the triangle is 180-125 = 55 because it is an adjacent supplementary angle to 125
y is the 3rd angle in the triangle so 69+55+y=180 angle sum of a triangle
$${\mathtt{69}}{\mathtt{\,\small\textbf+\,}}{\mathtt{55}}{\mathtt{\,\small\textbf+\,}}{\mathtt{y}} = {\mathtt{180}} \Rightarrow {\mathtt{y}} = {\mathtt{56}}$$
Now consider the big triangle. you can get the top angle because it is adjacent supplementary to 111
and now you can get x from angle sum of a triangle.
See if you can do the other 2 on your own but if you can't post back and I will help.
I might just do posts with the pics incase you need help since I have already copied them. :)
Finish this question totally but then I think that you should start a new post in your new name :)
In the little triangle on the right containing x, the angle at the top right = 40 degrees because it is equal to the opposite angle in the parallelogram formed by the intersection of the parallel transversals with the double and triple arrows. And the other angle in this triangle = 67 degrees. It's the corresponding angle to the other 67 degree angle formed by the intersection of parallel transversals with the single and triple arrows. Thus. x = 180 - 67 - 40 = 73 degrees.
And in the lttle triangle at the top....the angle at the bottom right is the vertical angle to the 67 degree angle in the triangle containing x. And the angle at the bottom left of this small triangle is the corresponding angle to the 40 degree angle formed by the intersection of the parallel transversals with the double and triple arrows. So y = 180 - 67 - 40 = 73 degrees, as well.
x is supplemental to the 56 degree angle = 124 degrees
And y is the same measure as the angle between the 41 and 56 degree angles at the top left. And these three angles sum to 180 degrees. So, y = 180 - 56 - 41 = 83 degrees.
Wow never knew it could be solved so easily when you just find the equal opposite of it and/or making sure it adds up to 180.
Thanks CPhill & Melody.
I will try this and hopefully get most of it right.
Well i got the first bit all right.
But the algebra bit i got 2 wrong.
But atleast i learnt something.
Also, Melody i will start a new post the second i got a new question.