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I often solve math questions because I like it (This may sound crazy, I know :)). Today I came across an interesting permutation & combination question.

The question is as follows:

6 people (named A, B, C, D, E, F) are in a line in a supermarket. In how many possible arrangements is C between A and B?

Options: A. 30 B. 60 C. 120 D. 240 (<- Correct answer) E. 360

The previous question was like: "In how many possible arrangements C is in front of A". My answer was: 6!/2 which is 360. Because in one half, C is in front of A, and in the other half A is in front of C. (No calculation required, common sense & interpretation & mathematical intuition is sufficient to find the correct answer I believe.) But this one really puzzled me. I will be grateful if you provide a clear explanation. Is there any way to solve this question without writing down all combinations? Any practical method? Because this question is from a math book which prepares students for the university entrance exam in my country in which every question should not take more than 30 sec to be solved.

Let me know if the questions is unclear (because I translated it, probably in a bad way)

Many many thanks in advance.

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    $\begingroup$ This turns out to be a good combinatorics problem for illustrating the importance of reading the stated condition carefully: C does not need to be immediately between A and B, only somewhere between them (as I painfully realized for my answer)... $\endgroup$ – colormegone May 3 '14 at 6:56
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Consider the ways to arrange $\{X, X, X, D, E, F\}$ where the X are placeholders --spots where A,B,C will stand so they are in the required order.

We don't care about the order of the X; there are $3!$ ways to arrange them but they are all the same arrangement. So the arrangements of six symbols can be divided into $\frac{6!}{3!}$ groups of equivalent arrangements.

Next consider that placing $C$ between $A$ and $B$ can be done in two ways. $ACB$ and $BCA$.

So the count you require of $\frac{2\times 6!}{3!} = 240$.

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All you care about is the order of $A,B,C$. There are $3!=6$ orders of them,of which $2$ have $C$ in the middle. So $\frac 26=\frac 13$ of the arrangements have $C$ in the middle. $\frac {6!}3=240$

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  • $\begingroup$ @ Ross Millikan , see this question is not asking about ACB or BCA, it may be ADCB or something else like that. I didn't get how that case is being taken care in your argument. Kindly explain. $\endgroup$ – Sry Mar 31 '15 at 9:51
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    $\begingroup$ @Sry: my answer is in the same spirit as Grahmam Kemp's. $A,B,C$ are sprinkled among the rest and all we care about is their relative order. Two of the six possible orders match what is asked. $\endgroup$ – Ross Millikan Mar 31 '15 at 14:17
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[This problem was making me crazy -- because I found three arguments for why the answer has to be 48 -- until I realized the statement doesn't mean C is immediately in between A and B...]

Here's another way to look at the arrangements. In order for C to be between A and B, there are only four places where that person can stand:

$$ \bullet \ C \ \bullet \ \bullet \ \bullet \ \bullet \ \ \ , $$ $$ \bullet \ \bullet \ C \ \bullet \ \bullet \ \bullet \ \ \ , $$ $$ \bullet \ \bullet \ \bullet \ C \ \bullet \ \bullet \ \ \ , $$ $$ \bullet \ \bullet \ \bullet \ \bullet \ C \ \bullet \ \ \ . $$

So there are two cases, one where C is one position away from either end, the other where they are two positions away.

In the first case, there is only one spot to stand available on one side of C , so we must have either A or B stand there. Whichever of the two does so, the other may take any of the remaining four positions on the other side of C, along with the other three people. Those four can then be arranged in any order and still met the requirement that C is between A and B:

$$ A \ C \quad 4! \ \ \ , $$ $$ B \ C \quad 4! \ \ \ ; $$

or

$$ 4! \quad C \ A \ \ \ , $$ $$ 4! \quad C \ B \ \ \ . $$

So this case accounts for $ \ 2 \ \cdot \ 2 \ \cdot \ 4! \ = \ 96 \ $ possible arrangements.

In the second case, there are two positions to one side of C and three positions to the other. Again, if A is on one side of C , B must be on the other. On the side of C with two positions available, we could have A or B standing there along with one of the three other people; they can take up one of two possible arrangements. To the other side, the remaining three people may be arranged in any order. We therefore may have:

$$ (A \ \bullet \ : \ \ 3 \cdot 2!) \quad C \quad 3! \ \ \ , $$ $$ (B \ \bullet \ : \ \ 3 \cdot 2!) \quad C \quad 3! \ \ \ ; $$

or

$$ 3! \quad C \quad (A \ \bullet \ : \ \ 3 \cdot 2!) \ \ \ , $$ $$ 3! \quad C \quad (B \ \bullet \ : \ \ 3 \cdot 2!) \ \ \ . $$

This second case permits $ \ 2 \ \cdot \ 2 \ \cdot \ 3 \ \cdot \ 2! \ \cdot \ 3! \ = \ 144 \ $ possible arrangments.

Hence, there are $ \ 96 \ + \ 144 \ = \ 240 \ $ ways in which these six people may stand in line, subject to the specified condition.

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EDIT (5/6) -- An alternative method would be to solve the complementary problem. There are 6! = 720 possible arrangments of the six people in the line. There are four situations in which person C does not find themselves standing between A and B :

I : C is at one end of the line, with the other 5 people arranged in any of 5! = 120 ways, thus

$$ C \quad 5! \quad \quad \text{or} \quad \quad 5! \quad C $$

for a total of 240 arrangements;

II : C is standing one position in from either end of the line, with some one of the three people other than A or B at that end,

$$ 3 \quad C \quad 4! \quad \quad \text{or} \quad \quad 4! \quad C \quad 3 $$

permitting $ \ 2 \ \cdot \ 3 \ \cdot \ 4! \ = \ 144 \ $ arrangements;

III : C is standing two positions in from either end of the line, with both A and B together on that end,

$$ (A \ B \ : \ \ 2!) \quad C \quad 3! \quad \quad \text{or} \quad \quad 3! \quad C \quad (A \ B \ : \ \ 2!) $$

giving us $ \ 2 \ \cdot \ 2! \ \cdot \ 3! \ = \ 24 \ $ arrangements; and

IV : C is standing two positions in from either end of the line, with any two of the three people other than A or B together on that end,

$$ (\bullet \ \bullet \ : \ \ _3P_2) \quad C \quad 3! \quad \quad \text{or} \quad \quad 3! \quad C \quad (\bullet \ \bullet \ : \ \ _3P_2) $$

for $ \ 2 \ \cdot \ 3 \cdot 2 \ \cdot \ 3! \ = \ 72 \ $ arrangements.

Removing these complementary possibilities again gives us

$$ \ 720 \ - \ ( \ 240 \ + \ 144 \ + \ 24 \ + \ 72 \ ) \ = \ 720 \ - \ 480 \ = \ 240 \ $$

arrangements for the requirement that C be standing somewhere between A and B.

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