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Arranging cats and dogs - what is wrong with my approach

Four dogs and five cats race. In how many ways can this occur?What's wrong with my permutation logic?Find the number of ways this can be arranged in which no 2 women and no 2 men sit together given 4 men and 3 women are seated in a dinner table?In how many ways can the letters of word $PERMUTATIONS$ be arranged if there are always 4 letters between P and S?What is wrong in my attempt in permutations?Story Of naive Cats And Machiavellian MonkeyArranging $A$'s and $B$'s.Permutations of finishing a raceHow many strings of $6$ digits are there which use only the digits $0, 1$, or $2$ and in which $2$, whenever it appears, always does so after $1$?In how many ways can $n$ dogs and $k$ cats be arranged in a row so that no two cats are adjacent?

7

1

$begingroup$

We have 4 dogs and 3 cats in a line but no two cats can be together, in how many ways can they be arranged?

Since there are 5 spaces the cats can be in with the dogs fixed, there are ${5 choose 3} * 4! * 3! = 1440$ ways and this is the correct answer.

I thought of a different approach. Instead of fixing the dogs' places, I fixed the places of the cats. Now, we have 4 spaces of which the two spaces in the middle must be filled. Therefore, out of the 4 dogs, 2 must fill those, and there are $4 * 3$ ways of doing this (since one dog must be chosen to fill one middle space and the other, to fill the second but now there are only 3 dogs left.)

The other two dogs are free to go to any of the 4 spaces, with $4^2$ possibilities.

The cats can now be arranged in $3!$ ways.

So, our final answer should be $3! * 4^2 * 4 * 3 = 1152$

Where have I gone wrong?

combinatorics permutations

share|cite|improve this question

asked Mar 30 at 11:26

Akshat AgarwalAkshat Agarwal

513

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Is your problem arising as a consequence of a rainfall ?
  $endgroup$
  – Jean Marie
  Mar 30 at 16:27
  

  
  
  
  

add a comment | 

7

1

$begingroup$

We have 4 dogs and 3 cats in a line but no two cats can be together, in how many ways can they be arranged?

Since there are 5 spaces the cats can be in with the dogs fixed, there are ${5 choose 3} * 4! * 3! = 1440$ ways and this is the correct answer.

I thought of a different approach. Instead of fixing the dogs' places, I fixed the places of the cats. Now, we have 4 spaces of which the two spaces in the middle must be filled. Therefore, out of the 4 dogs, 2 must fill those, and there are $4 * 3$ ways of doing this (since one dog must be chosen to fill one middle space and the other, to fill the second but now there are only 3 dogs left.)

The other two dogs are free to go to any of the 4 spaces, with $4^2$ possibilities.

The cats can now be arranged in $3!$ ways.

So, our final answer should be $3! * 4^2 * 4 * 3 = 1152$

Where have I gone wrong?

combinatorics permutations

share|cite|improve this question

asked Mar 30 at 11:26

Akshat AgarwalAkshat Agarwal

513

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Is your problem arising as a consequence of a rainfall ?
  $endgroup$
  – Jean Marie
  Mar 30 at 16:27
  

  
  
  
  

add a comment | 

$begingroup$

We have 4 dogs and 3 cats in a line but no two cats can be together, in how many ways can they be arranged?

Since there are 5 spaces the cats can be in with the dogs fixed, there are ${5 choose 3} * 4! * 3! = 1440$ ways and this is the correct answer.

I thought of a different approach. Instead of fixing the dogs' places, I fixed the places of the cats. Now, we have 4 spaces of which the two spaces in the middle must be filled. Therefore, out of the 4 dogs, 2 must fill those, and there are $4 * 3$ ways of doing this (since one dog must be chosen to fill one middle space and the other, to fill the second but now there are only 3 dogs left.)

The other two dogs are free to go to any of the 4 spaces, with $4^2$ possibilities.

The cats can now be arranged in $3!$ ways.

So, our final answer should be $3! * 4^2 * 4 * 3 = 1152$

Where have I gone wrong?

combinatorics permutations

share|cite|improve this question

asked Mar 30 at 11:26

Akshat AgarwalAkshat Agarwal

513

$endgroup$

share|cite|improve this question

asked Mar 30 at 11:26

Akshat AgarwalAkshat Agarwal

513

share|cite|improve this question

asked Mar 30 at 11:26

Akshat AgarwalAkshat Agarwal

513

asked Mar 30 at 11:26

Akshat AgarwalAkshat Agarwal

513

asked Mar 30 at 11:26

Akshat AgarwalAkshat Agarwal

513

2 Answers
2

active

oldest

votes

6

$begingroup$

The second computation is missing a symmetry. Say your initial pattern is $$underline {quad}C_1underline {quad}C_2underline {quad}C_3underline {quad}$$

You then populate the spaces immediately to the right of $C_1$, and $C_2$. As:

$$underline {quad}C_1D_1underline {quad}C_2D_2underline {quad}C_3underline {quad}$$

So far so good. You still have $D_3,D_4$ to place. Where can they go? True, they can each go to any of the four spaces, but if, say, they both go to the first space, in which order do they go?

Taking the two possible orders into account, we see that you are missing $$4times 3!times 4times 3=288$$ cases. Adding them back gives you the desired result.

Phrased differently: once you have placed $D_3$ there are now five available spaces for $D_4$ (since $D_4$ might go either to the left or to the right of $D_3$). thus you should have had $$3!times 4times 5times 4times 3=1440$$

share|cite|improve this answer

edited Mar 30 at 15:26

answered Mar 30 at 11:44

lulululu

43.6k25081

$endgroup$

add a comment | 

0

$begingroup$

You can separate the two cases for the two last dogs: single dogs and double dogs.

Single dogs:
$$P(4,2)=frac{4!}{2!}=12.$$
Double dogs:
$$P(2,2)cdot C(4,1)=2cdot 4=8.$$
Hence, there are $12+8=20$ (not $4^2=16$) ways to distribute the last two dogs.

The final answer is:
$$3!cdot 20cdot 4cdot 3=1440.$$

share|cite|improve this answer

answered Mar 30 at 15:27

farruhotafarruhota

21.8k2842

$endgroup$

add a comment | 

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6

$begingroup$

The second computation is missing a symmetry. Say your initial pattern is $$underline {quad}C_1underline {quad}C_2underline {quad}C_3underline {quad}$$

You then populate the spaces immediately to the right of $C_1$, and $C_2$. As:

$$underline {quad}C_1D_1underline {quad}C_2D_2underline {quad}C_3underline {quad}$$

So far so good. You still have $D_3,D_4$ to place. Where can they go? True, they can each go to any of the four spaces, but if, say, they both go to the first space, in which order do they go?

Taking the two possible orders into account, we see that you are missing $$4times 3!times 4times 3=288$$ cases. Adding them back gives you the desired result.

Phrased differently: once you have placed $D_3$ there are now five available spaces for $D_4$ (since $D_4$ might go either to the left or to the right of $D_3$). thus you should have had $$3!times 4times 5times 4times 3=1440$$

share|cite|improve this answer

edited Mar 30 at 15:26

answered Mar 30 at 11:44

lulululu

43.6k25081

$endgroup$

add a comment | 

6

$begingroup$

The second computation is missing a symmetry. Say your initial pattern is $$underline {quad}C_1underline {quad}C_2underline {quad}C_3underline {quad}$$

You then populate the spaces immediately to the right of $C_1$, and $C_2$. As:

$$underline {quad}C_1D_1underline {quad}C_2D_2underline {quad}C_3underline {quad}$$

So far so good. You still have $D_3,D_4$ to place. Where can they go? True, they can each go to any of the four spaces, but if, say, they both go to the first space, in which order do they go?

Taking the two possible orders into account, we see that you are missing $$4times 3!times 4times 3=288$$ cases. Adding them back gives you the desired result.

Phrased differently: once you have placed $D_3$ there are now five available spaces for $D_4$ (since $D_4$ might go either to the left or to the right of $D_3$). thus you should have had $$3!times 4times 5times 4times 3=1440$$

share|cite|improve this answer

edited Mar 30 at 15:26

answered Mar 30 at 11:44

lulululu

43.6k25081

$endgroup$

add a comment | 

$begingroup$

The second computation is missing a symmetry. Say your initial pattern is $$underline {quad}C_1underline {quad}C_2underline {quad}C_3underline {quad}$$

You then populate the spaces immediately to the right of $C_1$, and $C_2$. As:

$$underline {quad}C_1D_1underline {quad}C_2D_2underline {quad}C_3underline {quad}$$

So far so good. You still have $D_3,D_4$ to place. Where can they go? True, they can each go to any of the four spaces, but if, say, they both go to the first space, in which order do they go?

Taking the two possible orders into account, we see that you are missing $$4times 3!times 4times 3=288$$ cases. Adding them back gives you the desired result.

Phrased differently: once you have placed $D_3$ there are now five available spaces for $D_4$ (since $D_4$ might go either to the left or to the right of $D_3$). thus you should have had $$3!times 4times 5times 4times 3=1440$$

share|cite|improve this answer

edited Mar 30 at 15:26

answered Mar 30 at 11:44

lulululu

43.6k25081

$endgroup$

share|cite|improve this answer

edited Mar 30 at 15:26

answered Mar 30 at 11:44

lulululu

43.6k25081

answered Mar 30 at 11:44

lulululu

43.6k25081

answered Mar 30 at 11:44

lulululu

43.6k25081

0

$begingroup$

You can separate the two cases for the two last dogs: single dogs and double dogs.

Single dogs:
$$P(4,2)=frac{4!}{2!}=12.$$
Double dogs:
$$P(2,2)cdot C(4,1)=2cdot 4=8.$$
Hence, there are $12+8=20$ (not $4^2=16$) ways to distribute the last two dogs.

The final answer is:
$$3!cdot 20cdot 4cdot 3=1440.$$

share|cite|improve this answer

answered Mar 30 at 15:27

farruhotafarruhota

21.8k2842

$endgroup$

add a comment | 

0

$begingroup$

You can separate the two cases for the two last dogs: single dogs and double dogs.

Single dogs:
$$P(4,2)=frac{4!}{2!}=12.$$
Double dogs:
$$P(2,2)cdot C(4,1)=2cdot 4=8.$$
Hence, there are $12+8=20$ (not $4^2=16$) ways to distribute the last two dogs.

The final answer is:
$$3!cdot 20cdot 4cdot 3=1440.$$

share|cite|improve this answer

answered Mar 30 at 15:27

farruhotafarruhota

21.8k2842

$endgroup$

add a comment | 

$begingroup$

You can separate the two cases for the two last dogs: single dogs and double dogs.

Single dogs:
$$P(4,2)=frac{4!}{2!}=12.$$
Double dogs:
$$P(2,2)cdot C(4,1)=2cdot 4=8.$$
Hence, there are $12+8=20$ (not $4^2=16$) ways to distribute the last two dogs.

The final answer is:
$$3!cdot 20cdot 4cdot 3=1440.$$

share|cite|improve this answer

answered Mar 30 at 15:27

farruhotafarruhota

21.8k2842

$endgroup$

share|cite|improve this answer

answered Mar 30 at 15:27

farruhotafarruhota

21.8k2842

answered Mar 30 at 15:27

farruhotafarruhota

21.8k2842

answered Mar 30 at 15:27

farruhotafarruhota

21.8k2842