The rule says that A must come before B, and B must come before A.coolhabhi wrote:There are 6 persons - A, B, C, D, E and F. They are to be seated in a row such that B never sits anywhere ahead of A, and C never sits anywhere ahead of B. In how many different ways can this be done?
(a) 60
(b) 72
(c) 120
(d) 600
(e) 700
B never sits anywhere ahead of A, and C never sits anywhere ahead of B.coolhabhi wrote:There are 6 persons - A, B, C, D, E and F. They are to be seated in a row such that B never sits anywhere ahead of A, and C never sits anywhere ahead of B. In how many different ways can this be done?
(a) 60
(b) 72
(c) 120
(d) 600
(e) 700
GMATGuruNY wrote:B never sits anywhere ahead of A, and C never sits anywhere ahead of B.coolhabhi wrote:There are 6 persons - A, B, C, D, E and F. They are to be seated in a row such that B never sits anywhere ahead of A, and C never sits anywhere ahead of B. In how many different ways can this be done?
(a) 60
(b) 72
(c) 120
(d) 600
(e) 700
Implication:
A sits somewhere to the left of B, while B sits somewhere to the left of C.
Number of options for D = 6. (Any of the 6 seats.)
Number of options for E = 5. (Any of the 5 remaining seats.)
Number of options for F = 4. (Any of the 4 remaining seats.)
Number of options for A = 1. (Of the remaining 3 seats, the leftmost must be occupied by A.)
Number of options for B = 1. (Of the remaining 2 seats, the one on the left must be occupied by B.)
Number of options for C = 1. (Only 1 seat left.)
To combine these options, we multiply:
6*5*4*1*1*1 = 120.
The correct answer is C.
That starts off ok, and is based on a good concept, but this, 6!/3!, does not take into account the fact that A must be to the left of B, who must be to the left of C.Amrabdelnaby wrote:Why can't we do it in the following steps:
Stage 1: we will first seat A, B & C then seat the rest: 1x1x1x3x2x1
Stage 2: then we see how many possible ways to seat A, B & C in six chairs: 6!/3!
then we multiply both stages 1 & 2 together.
what is wrong exactly in this process?
GMATGuruNY wrote:B never sits anywhere ahead of A, and C never sits anywhere ahead of B.coolhabhi wrote:There are 6 persons - A, B, C, D, E and F. They are to be seated in a row such that B never sits anywhere ahead of A, and C never sits anywhere ahead of B. In how many different ways can this be done?
(a) 60
(b) 72
(c) 120
(d) 600
(e) 700
Implication:
A sits somewhere to the left of B, while B sits somewhere to the left of C.
Number of options for D = 6. (Any of the 6 seats.)
Number of options for E = 5. (Any of the 5 remaining seats.)
Number of options for F = 4. (Any of the 4 remaining seats.)
Number of options for A = 1. (Of the remaining 3 seats, the leftmost must be occupied by A.)
Number of options for B = 1. (Of the remaining 2 seats, the one on the left must be occupied by B.)
Number of options for C = 1. (Only 1 seat left.)
To combine these options, we multiply:
6*5*4*1*1*1 = 120.
The correct answer is C.
I'm assuming that your question is for Mitch and not me. Correct?Amrabdelnaby wrote:Hi Brent,
Thank you for the clear solution. However, I have a question.
Why did we not start by the most restrictive first? and in case we do here is how i thought about it.
please correct me if i am wrong.
_ _ _ _ _ _ we have six places and we want to sit the most
A B C 3 2 1
restrictive people first, A, B & C then we have 3 ways to sit the third, 2 ways to sit the second and one way to sit the third.
now we have 6 different ways to sit the 3 non restricted ones.
now if we take a look at the first 3 restricted ones, we should think of the possible ways to seat them in this form in 6 different chairs so we multiply the 6 we got earlier by 6C3 which is 20 to get 120.
Is this a correct thinking process?
Please advise
GMATGuruNY wrote:B never sits anywhere ahead of A, and C never sits anywhere ahead of B.coolhabhi wrote:There are 6 persons - A, B, C, D, E and F. They are to be seated in a row such that B never sits anywhere ahead of A, and C never sits anywhere ahead of B. In how many different ways can this be done?
(a) 60
(b) 72
(c) 120
(d) 600
(e) 700
Implication:
A sits somewhere to the left of B, while B sits somewhere to the left of C.
Number of options for D = 6. (Any of the 6 seats.)
Number of options for E = 5. (Any of the 5 remaining seats.)
Number of options for F = 4. (Any of the 4 remaining seats.)
Number of options for A = 1. (Of the remaining 3 seats, the leftmost must be occupied by A.)
Number of options for B = 1. (Of the remaining 2 seats, the one on the left must be occupied by B.)
Number of options for C = 1. (Only 1 seat left.)
To combine these options, we multiply:
6*5*4*1*1*1 = 120.
The correct answer is C.
