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Thread: Hard perms and combs question

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    Cadet JustRandomThings's Avatar
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    Hard perms and combs question

    Sorry for bombarding with questions but I really don't know how to approach this question

    Teach me the logic behind these questions pls Thanks in advance

    From the letters of the word WOOLLOOMOOLOO four letters are hosen are chosen at random
    Find the number of ways that
    (i) No O’s are chosen
    (ii) only O’s are chosen
    (iii) The three L’s are chosen
    (iv) no more than one Ois chosen
    (v) at least one L is chosen
    (vi) only O’s and L’s are chosen if it is known at least two L’s have been chosen
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    Re: Hard perms and combs question

    Quote Originally Posted by JustRandomThings View Post
    Sorry for bombarding with questions but I really don't know how to approach this question

    Teach me the logic behind these questions pls Thanks in advance

    From the letters of the word WOOLLOOMOOLOO four letters are hosen are chosen at random
    Find the number of ways that
    (i) No O’s are chosen
    (ii) only O’s are chosen
    (iii) The three L’s are chosen
    (iv) no more than one Ois chosen
    (v) at least one L is chosen
    (vi) only O’s and L’s are chosen if it is known at least two L’s have been chosen
    i) No O's so there's 5 letters and we must pick 4 - 5C4

    ii) 8 O's so 8C4

    iii) We pick the 3 L's through 3C3 then 1 letter from the others so 10C1, 3C3 x 10C1

    iv) Pick one O first, so 8C1, now pick the other 3 letters from the non-O's so 5C3 x 8C1

    v) For this we can do complementary event so total ways minus ways to pick zero L's. Total is just 13C4 and the ways to pick no L's is 10C4 as there are 3 L's so we can pick from 10 letters only. Therefore 13C4 - 10C4

    vi) Two L's are chosen at least so there's 2 or 3. For 2 it's 8C2 to pick the remaining O's. For 3 it's just 8C1 to pick the 1 O. Add these to get 8C2 + 8C1

    Correct me if I'm wrong, hope this helped.
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    Cadet JustRandomThings's Avatar
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    Re: Hard perms and combs question

    Quote Originally Posted by pikachu975 View Post
    i) No O's so there's 5 letters and we must pick 4 - 5C4

    ii) 8 O's so 8C4

    iii) We pick the 3 L's through 3C3 then 1 letter from the others so 10C1, 3C3 x 10C1

    iv) Pick one O first, so 8C1, now pick the other 3 letters from the non-O's so 5C3 x 8C1

    v) For this we can do complementary event so total ways minus ways to pick zero L's. Total is just 13C4 and the ways to pick no L's is 10C4 as there are 3 L's so we can pick from 10 letters only. Therefore 13C4 - 10C4

    vi) Two L's are chosen at least so there's 2 or 3. For 2 it's 8C2 to pick the remaining O's. For 3 it's just 8C1 to pick the 1 O. Add these to get 8C2 + 8C1

    Correct me if I'm wrong, hope this helped.
    For question iv) wouldn't you also include a case with no zeros?
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    Re: Hard perms and combs question

    Quote Originally Posted by JustRandomThings View Post
    For question iv) wouldn't you also include a case with no zeros?
    There are only 3 L's and it says only O's and L's are chosen so there must be an O.
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    Junior Member BenHowe's Avatar
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    Re: Hard perms and combs question

    You've missed some cases and arrangements. These letter ones are generally easier to do with perms I find
    Last edited by BenHowe; 10 Aug 2017 at 10:34 PM.
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    Re: Hard perms and combs question

    Quote Originally Posted by BenHowe View Post
    You've missed some cases and arrangements. These letter ones are generally easier to do with perms I find
    Yeah but idk about the wording since it just says the letters are chosen
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    Re: Hard perms and combs question

    For question 1, since you can only choose from 1 w , 1 M, 3 L and you need to choose 4, wouldn't there be only 3 cases. MLLL, WLLL, MWLL. The order is not important since the question states how many ways of choosing which implicates combination
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    Junior Member BenHowe's Avatar
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    Re: Hard perms and combs question

    well since it says chosen we are only interested in the total combinations not the arrangements of the words
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    Re: Hard perms and combs question

    Quote Originally Posted by JustRandomThings View Post
    For question 1, since you can only choose from 1 w , 1 M, 3 L and you need to choose 4, wouldn't there be only 3 cases. MLLL, WLLL, MWLL. The order is not important since the question states how many ways of choosing which implicates combination
    But there's still 5 ways u can pick the letters because the L's are different. You could pick WML1L2, WML1L3, WML2L3, WL1L2L3, ML1L2L3 which is 5C4 i.e. 5 ways to pick.
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    Junior Member BenHowe's Avatar
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    Re: Hard perms and combs question

    Quote Originally Posted by pikachu975 View Post
    But there's still 5 ways u can pick the letters because the L's are different. You could pick WML1L2, WML1L3, WML2L3, WL1L2L3, ML1L2L3 which is 5C4 i.e. 5 ways to pick.
    They dont explain that well in highschool. Because if you were asked to find the probability of one of the examples above you would need to regard the letters as distinguishable to produce equally likely outcomes, which you have inadvertently done. Sometimes it is valid to view the given letters as identical, it just depends on the situation
    pikachu975 likes this.
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    Re: Hard perms and combs question

    Quote Originally Posted by JustRandomThings View Post
    Sorry for bombarding with questions but I really don't know how to approach this question

    Teach me the logic behind these questions pls Thanks in advance

    From the letters of the word WOOLLOOMOOLOO four letters are hosen are chosen at random
    Find the number of ways that
    (i) No O’s are chosen
    (ii) only O’s are chosen
    (iii) The three L’s are chosen
    (iv) no more than one Ois chosen
    (v) at least one L is chosen
    (vi) only O’s and L’s are chosen if it is known at least two L’s have been chosen
    Hey, do you have answers?
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    Junior Member BenHowe's Avatar
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    Re: Hard perms and combs question

    Start of by stating, we regard the letters as distinguishable. If it were probability then you would say, to produce equally likely outcomes. But this is beyond the level at school.

    So we have 1W,8O,3L,1M, such that all the O's,L's are different. You can introduce subscript etc.

    i) If no O's are chosen we need to remove the 8O's from our total letters. We can now select any 4 letters from the total 5, in 5C4=10 ways

    ii) Similar to above our total letters now consist only of the 8O's. We can select any 4O's from the 8 in 8C4=70 ways

    iii) Now there is one possible selection for the 3L's i.e. L1L2L3. This can be done in 3C3=1 ways. We can then select 3 more letters from any of the 10 remaining letters in 10C1 ways. We can then put these two answers together by the multiplication rule (since if we selected the non-L first this would've have changed our ability to select the L's) in 3C3 x 10C1 = 10 ways

    iv) So we can do this two ways, we can look at the cases for <=1O or do a subtraction from the total unrestricted selection with the cases for O>=2

    Method 1: O<=1

    If we want no O's this is just our answer from (a) i.e. 5C4

    If we want 1O, we can select any of the O's from the total 8 in 8C1 ways. We can then select any 3 letters from the non-O's in 5C3 ways.. Putting this together with multiplication rule we get 8C1 x 5C3

    Thus the total is 5C4+8C1 x 5C3=85

    Method 2: If we can select 4 letters from any of the 13, we can do thuis in 13C4 ways

    If we want 2O's, we can select them in 8C2 ways and then the non-O's in 5C2 ways, =8C2 x 5C2

    If we want 3O's, we can select them in 8C3 ways and then the non-O's in 5C1 ways, =8C3 x 5C1

    If we want 4O's, we can select them in 8C4 ways and we dont select any non-O's

    Required answer = 13C4-(8C2 x 5C2+8C3 x 5C1+8C4)=85

    v) There is a kinda trick to this sort of question, you can do case by case but it is much easier to simply satisfy the initial condition and then go from there.

    We need at least one L to be chosen, we can select this L in 3C1 ways. We can then take an unrestricted selection of 3 letters from the remaining 12 letters, since regardless of the outcome we satisfy the initial condition i.e. L LLL or even L OOO etc. This can be done in 12C3 ways.

    Thus the required answer is 3C1 x 12C3=660

    Yeah if you do by cases, you can do it in a similar way to (iv) but you have to be careful of double counting

    vi)

    This can be done by cases

    Case 1: 2L | 20, the 2 O's can be selected from any of the 8 O's in 8C2 ways
    Case 2: 2L | 1O,1L the 1O can be selected from any of the 8 O's in 8C2 ways and the 1 L in 1C1 ways, so 8C2 x 1C1 =8C2
    Case 3: 3L | 1O ... 8C1
    Case 4: 3L | 1L but this is not valid since no O would be selected

    Thus the required answer =8C2+8C1+8C1=44

    Hope this helps. Feel free to ask any q
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    Junior Member BenHowe's Avatar
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    Re: Hard perms and combs question

    What I wrote was wrong because my initial assumption was incorrect. We must regard the same letters as identical. You only need to worry about outcomes when doing probability. Although don't worry about this for high school.
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