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 ?

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.

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..

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.

Once we fixed three positions for the family, then we placed son, father, grandfather in order in those psition s. Automatically no father is ahead of his son.

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

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.

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 ?

OOPS, got it !

Thank u anyway 🙂

Do keep the articles pouring in 🙂

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

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..

The father also is the son of the grandfather 🙂 Think about it…

regards

J

dammit.. night time is not the ideal time to do logic stuff.. 😐

if “no father wants to stand before his son” then what should i do

sorry got it

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

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?

Sir,

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

Yes it will

regards

J

Once we fixed three positions for the family, then we placed son, father, grandfather in order in those psition s. Automatically no father is ahead of his son.

regards

J

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?

Yes, effectively it is equivalent to the second method I mentioned!

regards

J

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