Find the number of arrangements of all the letters in the word EQUATION if the consonants are seperated?
Total no. of arrangements= 8!
No of arrangements in which consonants are tog = 6! X 3!
Ans = 8! - 6!X3! = 5760
is my ans right?
Originally posted by quailmaster:Find the number of arrangements of all the letters in the word EQUATION if the consonants are seperated?
Total no. of arrangements= 8!
No of arrangements in which consonants are tog = 6! X 3!
Ans = 8! - 6!X3! = 5760
is my ans right?
Hi quailmaster,
6! X 3! is the number of arrangements where all three consonants are together. You will also need to consider situations where only 2 of the consonants are together.
Cheers.
to arrange 2 consonants together i use 3X2 to choose the 2 consonants that are tog.. then use 6P2 to arrange them between the vowels
Originally posted by quailmaster:to arrange 2 consonants together i use 3X2 to choose the 2 consonants that are tog.. then use 6P2 to arrange them between the vowels
Hi quailmaster,
I just realise that 5760 is indeed the correct answer, though you are missing something in your workings. I wasn't a Maths student, so please do pardon me if my workings are a bit rough.
Number of arrangements where all three consonants are together
= 3! X 6 X 5! = 4320
Number of arrangements where two consonants are together
= (3 X 2) X 7 X 6! = 30240
Total number of arrangements with consonants separated
= 8! - 4320 - 30240 = 5760
Originally posted by TrueHeart:Hi quailmaster,
I just realise that 5760 is indeed the correct answer, though you are missing something in your workings. I wasn't a Maths student, so please do pardon me if my workings are a bit rough.
Number of arrangements where all three consonants are together
= 3! X 6 X 5! = 4320
Number of arrangements where two consonants are together
= (3 X 2) X 7 X 6! = 30240
Total number of arrangements with consonants separated
= 8! - 4320 - 30240 = 5760
actually its not necessary to get the answers for No. of arrangements where 3/2 consonants tgt...juz show the statement (......=3! X 6 X 5!) enuf....coz there may b cases where the answer for one might be 72398273948 xD so juz write out the equation can liao
Originally posted by TrueHeart:Hi quailmaster,
I just realise that 5760 is indeed the correct answer, though you are missing something in your workings. I wasn't a Maths student, so please do pardon me if my workings are a bit rough.
Number of arrangements where all three consonants are together
= 3! X 6 X 5! = 4320
Number of arrangements where two consonants are together
= (3 X 2) X 7 X 6! = 30240
Total number of arrangements with consonants separated
= 8! - 4320 - 30240 = 5760
There are 5 consonants and 3 vowels in the word: equation.
So ans = 8! - 6!3! = 5760
Originally posted by HyuugaNeji:There are 5 consonants and 3 vowels in the word: equation.
So ans = 8! - 6!3! = 5760
8!= 40320
6!3! =4320
how did 8!-6!3! become 5760 0_o
Originally posted by HyuugaNeji:There are 5 consonants and 3 vowels in the word: equation.
So ans = 8! - 6!3! = 5760
Hi HyuugaNeji,
Please do correct me if I am wrong, but I believe that there are 3 consonants ( N, Q, T ) and 5 vowels ( A, E, I, O, U ) instead of the other way round.
Cheers.
oops sorry i was wrong. Terribly wrong. Your answer is correct.
why cant we consider the direct method of 5! X 6C3X 3! ? Altho the ans is wrong.....
Originally posted by Z een:why cant we consider the direct method of 5! X 6C3X 3! ? Altho the ans is wrong.....
I agree. I did that way on the first try as well. The answer is 14,400.
Im pretty sure it's right, slot-in method is always reliable to use :)
There are 3 consonants, 5 vowels.
Firstly, 5! for number of ways to arrange the VOWELS.
Secondly, 6C3 for choosing the number of ways that you can slot the 3 consonants IN BETWEEN the consonants.
Thirdly, 3! for the switching of places between the 3 consonants slotted in.
Then tada you get your answer of 14400. :D
Originally posted by Wheref:I agree. I did that way on the first try as well. The answer is 14,400.
Im pretty sure it's right, slot-in method is always reliable to use :)
There are 3 consonants, 5 vowels.
Firstly, 5! for number of ways to arrange the VOWELS.
Secondly, 6C3 for choosing the number of ways that you can slot the 3 consonants IN BETWEEN the consonants.
Thirdly, 3! for the switching of places between the 3 consonants slotted in.
Then tada you get your answer of 14400. :D
Hi Wheref,
You must also consider the situation where 2 consonants are together, whereas the third consonant is separated from them.
Cheers.
Originally posted by TrueHeart:Hi Wheref,
You must also consider the situation where 2 consonants are together, whereas the third consonant is separated from them.
Cheers.
Hi everyone,
I just realise that my method is flawed as well, my humble apologies.
(3 X 2) X 7 X 6! also include the sets of 3 consonants being together.
Honestly speaking, I am not too sure now, but perhaps we could just use the equation 8! - (3 X 2) x 7 x 6!, since it includes the situations where all three consonants are together as well.
Cheers.
- i think 14400 is the correct answer. u have 5 consonants, hence 6 spaces to slot. therefore answer is 5! x 6C3 x 3! or 5! x 6P3 = 14400. this is the forward method.
- to use the backward method...
no. of ways = 8! - all the cases where there are consonants are together
this 1, better not use 7!*2!*3C2 (explaining later) to find the case whereby 2 consonants are together. to find the case whereby 2 consonants are together, do 5! * 6 * 2! * 5 * 3C2. basically what this means is u arrange EUAIO vowels first, den slot in any 2 of the consonants into the 6 spaces, den slot in the last consonant into 5 available spaces. and there are 3C2 ways of choosing the 2 consonants. this method will ensure that it dosent overlap with the case of 3 consonants together.
for the case of 3 consonants together, it is simply 6!3! = 4320.
since for the 2 cases there are no intersect, so we can minus off directly, number of ways = 8! - (5!* 6 * 2! * 5 * 3C2) - 6!3!
=8! - 21600 - 4320
= 14400.
i was originally using 7! * 2! * 3C2 after reading trueheart's first working, and i thought he didnt consider the case of the intersections, thats why its wrong. but after further thought, using 7! * 2! * 3C2 will mean that the case of 3 consonants together is a full subset of it. then 8! - 7! * 2! * 3C2 will seem logical, but the answer lacks by 4320 which is equal to 6!3!. i cannot figure out where the overcount is...hmm...
Hi,
When we do P&C, we aim to do the question using the most direct method.
There are 5 vowels and 3 consonants.
First, we place the 5 vowels. That can be done in 5! ways.
Next, we insert 3 consonants in between vowels in 6P3 ways.
Finally, we use the multiplication principle to arrive at 14400 ways.
The exclusion principle may be complicated at times.
Thanks!
Cheers,
Wen Shih
Hi,
Below is a quick summary of P&C:
0. Permutation is Combination with order being considered, since we see their connection in the following result: nPr = nCr x r!,
from which we see the multiplication principle of choosing, then ordering.
1. Addition principle (useful when there are several mutually exclusive cases).
2. Multiplication principle (useful when there are several actions to be taken in sequence).
3. Exclusion principle (useful when we have a case like this:
number of ways w/o restriction - number of ways in which two people are together
to find the number of ways in which two people are separated).
4. Principle of applying the restriction first (useful when we have a case like this:
first and last digit is odd,
in which we find the number of ways for that first).
5. Principle of handling repetitions (which usually involves division by the factorial of the number of identical items).
6. P&C is closely related to the topic of probability in the following generic way:
P(event A occuring) = (number of ways in which event A can occur) / (number of ways without restriction).
Thanks!
Cheers,
Wen Shih
Originally posted by wee_ws:Hi,
Below is a quick summary of P&C:
0. Permutation is Combination with order being considered, since we see their connection in the following result: nPr = nCr x r!,
from which we see the multiplication principle of choosing, then ordering.
1. Addition principle (useful when there are several mutually exclusive cases).
2. Multiplication principle (useful when there are several actions to be taken in sequence).
3. Exclusion principle (useful when we have a case like this:
number of ways w/o restriction - number of ways in which two people are together
to find the number of ways in which two people are separated).
4. Principle of applying the restriction first (useful when we have a case like this:
first and last digit is odd,
in which we find the number of ways for that first).
5. Principle of handling repetitions (which usually involves division by the factorial of the number of identical items).
6. P&C is closely related to the topic of probability in the following generic way:
P(event A occuring) = (number of ways in which event A can occur) / (number of ways without restriction).
Thanks!
Cheers,
Wen Shih
Hi Dr Wee,
Thank you for your kind explanation. Once again, I apologise here for misleading everybody.
On a side note, I am not a "he".
Cheers.
Hi TrueHeart,
Thanks for the title, for which I am much flattered :) I hope to earn that in a few years' time. I'm always happy for anyone to call me Wen Shih.
No worry, for we are all here to learn and to make progress in our understanding of the world of mathematics and about life :)
Keep your contributions coming, for the benefit of all. Have a nice day!
Cheers,
Wen Shih
@trueheart
OT a bit, i saw that u said u werent a maths student, meaning that u nvr take maths during jc?
Originally posted by absol:@trueheart
OT a bit, i saw that u said u werent a maths student, meaning that u nvr take maths during jc?
Hi absol,
I studied in JC under the older three subjects syllabus, thus I did not need to take a contrast subject as current JC students need to. I did study A. Maths during my secondary school time though.
Cheers.
Lol life is good in the past :( Only 3 subjects to study
Originally posted by Wheref:Lol life is good in the past :( Only 3 subjects to study
Sorry but you are mistaken about only 3 subjects. It is 7. 4 A level subjects plus 3 AO level subjects( if you didn't take A maths in sec school as a pass in O level AM or AO level Maths is required for most uni courses).
And we didn't have graphical calculator to do the thinking for us. And the sci calculator can't solve simultaneous, quad and cubic equation for us.
Might I also mention that today's AM is a watered down version... man you guys have it really relaxed nowadays... learnt full kinematics with gravity and resolutions of forces and statistics with P&C, conditional probability, confidence interval testing etc etc in the good old early 90s... now all these are at H2 maths level instead.
Life is good today. All the watered down syllubuses...
I find that students rely too much on graphic calculators nowadays.... Being trained in the old syllabus, there are many questions in which I could see the solutions without the graphs even being plotted out on the graphic calculator...
We were trained to have a feel of graphs back then.
Hi,
In my humble opinion, the maths syllabuses of today lack rigour and depth.
To quote an example, students only need to know how to use GC to solve a system of linear equations without a solid understanding of linear algebra concepts.
Thanks!
Cheers,
Wen Shih