19 thoughts on “PnC Examples 3

  1. In the last question, it’s mentioned that no father wants to stand in front of his son right?
    So for any trio of father,paternal grandfather and son, we can have an arrangement as,
    FSP_ _ _ _ _ _ , or FPS_ _ _ _ _ _ right?
    So, don’t we have to multiply each nCr term with 2 ?

  2. When u divide 9 into 3 equal groups where order is not important… i guess u need to divide by one more 3! which signifies order is not important… because the groups of three can be interchanged among themselves… so it will be 1680/3! = 280…
    please clarify this whole concept of order as soon as possible…. the blog is really great… lets do away with the these little
    obscurities as much as we can

    • Girish, that will happen if the *groups* in question are identical (a situation I will address in a later post). But here, the groups are distinct; the families of P, Q and R. Only the position *within* a group is rendered irrelevant as internal order is already fixed as son, father, grandfather. So, only for each of the groups there will be a 3! in the denominator.

      regards
      J

  3. sir dbt : amusement park wala Q.. first we selct 3 pos. for 1st family in 9C3 ways,now no father stands before son,of the total ways of arranging 3 members of a family,i.e. 3! in 3 cases son will be ahead.. didnt understand only one to arrange..

  4. Sir,

    One doubt in the last question:
    We have divided 9 people in groups of 3 i.e. p,q,r in 1680 ways
    but these 3 families can be arranged in 6 ways, should not we consider this in the final answer?

    • No Gaurav, we have already picked three specific places for P’s family, 3 others for Q’s and the rest for R’s. So the arrangement within families is already taken care of.

      regards
      J

  5. Sir,
    In the first question, the same logic will be applicable to all the letters irrelevant of its sequence.
    it could be A & C or F & B ?
    wont it?

  6. Sir,
    in the last question, while following first method, how did you take care of the condition of father not standing ahead of son?

  7. Hi for the last question i did in this fashion
    So total arrangements possible are 9!.
    So first family arrangement it will be (1/6*9!)
    Second family arrangement it will be [1/6(1/6*9!)]
    Third family arrangement will be 1/6[1/6(1/6*9!)] =1680 ways

    Is my way of approach correct please explain?

  8. Sir, In the 2nd Question,
    while arranging A,B,C,D,E and F, BEC is the only order that satisfies the condition. So can’t we group B-E-C as one and since we are left with other 3: A,D,F, we’ll only have to arrange these 4, which gives us 4! = 24?
    Am I wrong?

    • Your answer would be right if BEC need to be consecutive! But B only needs to be somewhere before E (not immediately in front of him) and E somewhere before C.

      regards
      J

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