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Permutations & Combinations - 6 Persons

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by coolhabhi » Sat Aug 08, 2015 12:45 pm
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
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by Brent@GMATPrepNow » Sat Aug 08, 2015 1:27 pm
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
The rule says that A must come before B, and B must come before A.
So, the arrangement DAEBCF is good, and the arrangement CDAEBF is not good.

Okay, let's first IGNORE the rule about who can sit where.
If we ignore the rule, we can arrange the 6 people in 6! ways (720 ways)

Now let's examine one arrangement: AEDCBF (this is a bad arrangement)
Now, if we keep the D, E and F in their same places, in how many ways can we move the A, B and C around?
Well, there are 3 letters, so we can arrange them in 3! ways (6 ways).
They are:
1) AEDCBF - bad arrangement
2) AEDBCF - GOOD arrangement
3) BEDACF - bad arrangement
4) BEDCAF - bad arrangement
5) CEDABF - bad arrangement
6) CEDBAF - bad arrangement
Notice that, of the 6 arrangements, only 1 follows the given rule.
In other words, 1/6 of the 720 arrangements will follow the given rule.

1/6 of 720 = [spoiler]120 = C[/spoiler]


Here's a similar question: https://www.beatthegmat.com/mobster-comb ... 66632.html

Cheers,
Brent
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by GMATGuruNY » Sat Aug 08, 2015 1:54 pm
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.
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.
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by Amrabdelnaby » Sun Nov 15, 2015 5:58 am
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:
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.
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.
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by MartyMurray » Sun Nov 15, 2015 6:10 am
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?
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.
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by Amrabdelnaby » Wed Nov 18, 2015 11:07 am
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:
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.
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.
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by Brent@GMATPrepNow » Wed Nov 18, 2015 1:06 pm
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:
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.
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?

Cheers,
Brent
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