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Help with Statistics Question
- Thread starter Zizu1992
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if ur answer comes out correct , then u r right!Zizu1992 said:
there is more than one way to solve a question!!! marking scheme may differ but the thing that really really matters is the answer that u get and that the procedure u have carried out is correct!
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english please!Zizu1992 said:Us say nahi ata
then this is the wrong way !
this might help
http://www.mathsisfun.com/combinatorics ... tions.html
if this doesnt help ,plz type the question here!ill solve it
see, 9P9 gives you the number of ways the letters can be arranged, which is very good, but the question asks for "how many different arrangements" are possible. so the point is, there are two Ds and two Ls. the result with 9P9 counts the same arrangement with one L, and then the other L, but if you write out the arrangement, it's the same right? (<-- this sounds confusing; the next paragraph'll help more i think)
okay, i'll try to explain it this way: the letters are GOLD MEDAL, there are very many ways you can arrange them including this arrangement: GOLD MEDAL. so the permutation says they're two different things, but they aren't!
so to get rid of the "duplicates" we take the total number of ways they can be arranged (= 9!) and divide it by the (number of times each letter is repeated)'s permutation, in this case = 2! 2! (all the others except for L and D aren't repeated) giving u, in simple form, (9!)/(2! 2!) which gives the right answer...
okay, i'll try to explain it this way: the letters are GOLD MEDAL, there are very many ways you can arrange them including this arrangement: GOLD MEDAL. so the permutation says they're two different things, but they aren't!
so to get rid of the "duplicates" we take the total number of ways they can be arranged (= 9!) and divide it by the (number of times each letter is repeated)'s permutation, in this case = 2! 2! (all the others except for L and D aren't repeated) giving u, in simple form, (9!)/(2! 2!) which gives the right answer...
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amna mashallah ur good in dis.
how do u learn dis????
i really find these questions hard.
for da word: NECESSARY
Q1 HOW many seletions r there for 4 letters?
q2 how many arrangements are there of four letters?
how do u learn dis????
i really find these questions hard.
for da word: NECESSARY
Q1 HOW many seletions r there for 4 letters?
q2 how many arrangements are there of four letters?
- Messages
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ANYONE!!!!!!!!!!!!!!!
- Messages
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THANK U 4 TRYING
but i think thts definitely wrong
but i think thts definitely wrong
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anyone???
actually for the first one, if u do just 9C4, it doesn't give you the right answer... you have to assume a lot of different case and all, i tried doing the same kind of question with "possesses", didn't succeed, but when i looked at the solution it was really really weird. the good news, though, is that according to the syllabus all we're required to do is "solve simple problems involving selections" like selecting a team of 11 people out of 15 people, and the like. so you don't need to wreck your brains for (i). for permutations, we're required to "solve problems about arrangements of objects in a line including those involving repetition (e.g. number of ways of arranging the letters in the word NEEDLESS) or restriction (e.g. number of ways several people can stand next to each other if 2 of them must/must not stand next to each other)" so you don't need to wreck your brains for (ii) also if you look in the past papers, there are no questions like that... cheers!
also if you look in the past papers, there are no questions like that... cheers!
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thanx mate!
@Amina and @Ahmed. if u look at 2010 p6 (all variants) ull find that there are questions more serious and difficult than this one. You have to more than wreck ur brains over them. i dunno if u were talking about cambridge AS math p6, cuz u seem very relaxed. and the permutations and combinations questions are getting very diificult.
for example:
(i) Find the number of different ways that a set of 10 different mugs can be shared between Lucy
and Monica if each receives an odd number of mugs. [3]
(ii) Another set consists of 6 plastic mugs each of a different design and 3 china mugs each of a
different design. Find in how many ways these 9 mugs can be arranged in a row if the china
mugs are all separated from each other. [3]
(iii) Another set consists of 3 identical red mugs, 4 identical blue mugs and 7 identical yellow mugs.
These 14 mugs are placed in a row. Find how many different arrangements of the colours are
possible if the red mugs are kept together. [3]
This is a question from may 2010 p61.
for example:
(i) Find the number of different ways that a set of 10 different mugs can be shared between Lucy
and Monica if each receives an odd number of mugs. [3]
(ii) Another set consists of 6 plastic mugs each of a different design and 3 china mugs each of a
different design. Find in how many ways these 9 mugs can be arranged in a row if the china
mugs are all separated from each other. [3]
(iii) Another set consists of 3 identical red mugs, 4 identical blue mugs and 7 identical yellow mugs.
These 14 mugs are placed in a row. Find how many different arrangements of the colours are
possible if the red mugs are kept together. [3]
This is a question from may 2010 p61.