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AS permutation & combination questions .help required!

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1) A word is made out of 7 letters of EXAMPLE n some order. how many diff. words are possible if, the two letters E are next to each other?

2) 3 married couples , Mr Mrs lee, Mr Mrs martin, Mr Mrs shah stand in line for a photograph to be taken , find the diff. ways in which these six ppl can be arranged if (i) each man stands next to his wife (ii) no man stands next to his wife .

3) June 02/Q5(iii)
4)june03/ Q5(iii)
5) june05/Q7/b(ii)

please answer with some details because I'm having too many problems in s1 and specially this topic! thanks for your co-operation! :)
 
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1) The word EXAMPLE contains 7 letters(objects). Take the 2 'E's as one object since they are together. Therefore there are 6 objects now. The answer is 6! = 720 because the 2 'E's taken as one object cannot be interchanged.

2) (i) The diff. ways in which these six ppl can be arranged if each man stands next to his wife
There are 3 couples implies it is 3! ,but each couple contains a man and a woman, then the answer is 3! x 2! = 12

(ii) No man stands next to his wife ---> Total no of ways of arranging the 6 ppl LESS no of ways if each man stands next to his wife
6! - 12 = 708

For the other ques. consult the mark schemes. But if help is still needed, post again!
 
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anyways, thank you helper ..
i'm almost done with the past papers of p1 and s1 ..can anyone of you please guide me that what should i do next to to have a firm position to secure an A grade ..
 
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Relax. Read a couple of examiner reports again. Attempt some qsts which you find to trick you the most.

For maths the recipe for getting an A is simple: Understand the concept, give more weight to papers rather than book qsts, read examiner reports. And go with a fresh mind (means you sleep well the night before the exam).
 
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M stuk in permutation combinations too...hw cn i prepare apart frm s1 an d past papers...know any gud online math techer??
 
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S1 and Past Papers are the only way to prepare S1. If 'apart' means any other p1/p2/p3/m1/m2 then lemme knw of any specific topic here. I'll see to it.
 
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for q2, ii,
no man stands nxt to his wife,
M M M W W W

so,
no of ways to arrange the man, 3!
no of ways to arrange the couples, 4!

3! x 4! = 144

i not sure about the method thou, but i'll ask my math lecturer later.

good luck! ;)
 
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