Permutaions and combinations

Aishwarya Krishnaswamy · 2011-05-14 01:58:20

Hi I have couple doubts in the above topic.
1. In a crossword puzzle 20 words are to be guessed of which 8 words have each an alternative solution. What is the number of possible solution?
2. Find the number of ways in which an arrangement of 4 letters can be made from the word "Mathematics"
Thank You

Regards
Aishwarya

bobbym · 2011-05-14 02:16:20

Hi;

2) Use the exponential generating function.

Check the coefficient of x^4. It is 409 / 4. Now times that by 24 to get 2454. So there are 2454 arrangements of 4 letters.

gAr · 2011-05-14 02:25:36

Hi Aishwarya,

1)

Hi bobbym,

How are you today?

bobbym · 2011-05-14 02:35:29

Hi gAr;

Still feeling a little drained. But I am working on a nice problem and should post it soon.

How are you?

gAr · 2011-05-14 02:45:45

Ok, mathematics is the cure to all ills! Good to hear that you are working on a problem.

I'm fine.
Sometimes I find myself sitting in ergonomically awkward positions, so trying to avoid that.

bobbym · 2011-05-14 02:59:13

Actually the problem is working on me. I need to solve some smaller problems before I can even touch it.

I get up and walk around, go outside get some fresh air. Helps to focus your eyes on something other than the screen.

gAr · 2011-05-14 03:03:17

Ok.

Yes, that's good.

Aishwarya Krishnaswamy · 2011-05-15 00:22:20

Thank you very much to both of you. In the 1st question you had answered it as 2^8 ways. If going by the same logic then, the answer to this question - In how many ways can 9 letters be posted in 4 letter boxes - will be 4^9.

Could you please clarify if I am correct?

Also, I have questions.
1. A computer has 5 terminals and each terminal is capable of four distinct positions including the positions of rest what is the total number of signals that can be made?

2. A family comprised of an old man, 6 adults and 4 children is to be seated in a row with the condition that the children would occupy both the ends and nver occupy either side of the old man. How many sitting aarangements are possible?

3. A party of 6 is to be formed from 10 men and 7 women so as to include 3 men and 3 women. In how many ways can it be formed if 2 particular women refuse to join it?

Regards
Aishwarya

bobbym · 2011-05-15 00:38:04

Hi;

For this question:

In how many ways can 9 letters be posted in 4 letter boxes - will be 4^9.

It is not that simple. What are the letters?

gAr · 2011-05-15 02:15:44

Hi,

In how many ways can 9 letters be posted in 4 letter boxes

Let us consider all the possibilities,

Aishwarya Krishnaswamy · 2011-05-15 02:30:26

Thanks a lot for the help!!!

gAr · 2011-05-15 02:32:23

Hi,

Can you do the rest now?

Aishwarya Krishnaswamy · 2011-05-15 03:22:21

Could you help me with the 2nd question alone?

gAr · 2011-05-15 04:11:10

Hi Aishwarya,

I assume the condition like this: 2 children will occupy the ends, and other two children can sit anywhere except besides the old man.
I'm getting this answer: 4*3 * 7 * 6*5 * 6! = 1814400

But I'd like to know your answer too.

soroban · 2011-05-15 07:32:47

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Aishwarya Krishnaswamy · 2011-05-15 17:41:39

A family comprised of an old man, 6 adults and 4 children is to be seated in a row with the condition that the children would occupy both the ends and nver occupy either side of the old man. How many sitting aarangements are possible?

Hi,

Yes, I got the same answer 4*3 * 7 * 6*5 * 6!
Going by the logic that 2 children occupy the ends - 4*3,
then the selection of 2 adults next to the children - 6*5
The Old man has 7 options now to seat himself - *7
The balance 6 people ( 2 Children and 4 Adults) - 6!

But there is an anomaly in thinking like this. Since, in the 7 positions that the old man may occupy they are positions where he may have a child on his right or left or both which must be ruled out.

However, the source from where I picked this question, the answer is 4!*5*7!

gAr · 2011-05-15 19:17:55

Hi,

I did like this:
First fill the 2 ends: 4*3
Next, fill in the old man: 1
Besides the old man, 2 adults: 6*5
Now there are 6 people to be filled: 6!
Multiply the result by 7, considering the possible seats for the old man.
Hence the answer is 4*3 * 6*5 * 6! * 7

Another way of thinking the problem to be is:
Only children will be at the ends: that is one child at one end, 3 at the other. Or 2 children at both ends. : 4!*3
Fill in the old man: 5
Fill in the adults : 6!
Hence, the total = 4!*3 * 5 * 6!
= 6! * 5! * 3

But again, not the answer provided in the source.

Aishwarya Krishnaswamy · 2011-05-15 19:58:17

Hi,

Thanks for your help here.
The answer in the source is 4!*5!*7!. However, if we sum up the possibilities as we listed above this is what we get - 4!*6!*5*3+7!*5!*2.

Regards
Aishwarya

bobbym · 2011-05-15 20:02:56

Hi;

I am getting 1 814 400 by direct count. That agrees with the above answers and not with the source!

gAr · 2011-05-15 20:05:04

Hi Aishwarya,

You're welcome.
I do not understand 7!*5!*2 in your answer.

Aishwarya Krishnaswamy · 2011-05-15 20:35:06

I am sorry. It shoud be 7!*5!*3 and not 7!*5!*2. It was a typo. Its the simplification of 4*3 * 7 * 6*5 * 6!.

gAr · 2011-05-15 20:52:05

In post #16, you said the answer in the source is 4!*5*7!
And in #18, 4!*5!*7!

And the answer we get is 3*5!*7!

Shall we deduce that the source has a mistake in its answer?
Maybe it actually meant 4!*15*7!

Aishwarya Krishnaswamy · 2011-05-15 22:47:13

Yes I think we sould be concluding that the answer in the source is incorrect. It was good cracking it. Thanks.

gAr · 2011-05-15 22:56:55

You're welcome.

Aishwarya Krishnaswamy · 2011-05-25 20:16:42

Hello everybody,

I am back with a new problem in P&C

Q - There are stalls for 10 animals in a ship. In how ways can the shipload be made if there cows, calves and horses, and there are not less than 10 animals of each kind?

Regards
Aishwarya

Math Is Fun Forum

#1 2011-05-14 01:58:20

Permutaions and combinations

#2 2011-05-14 02:16:20

Re: Permutaions and combinations

#3 2011-05-14 02:25:36

Re: Permutaions and combinations

#4 2011-05-14 02:35:29

Re: Permutaions and combinations

#5 2011-05-14 02:45:45

Re: Permutaions and combinations

#6 2011-05-14 02:59:13

Re: Permutaions and combinations

#7 2011-05-14 03:03:17

Re: Permutaions and combinations

#8 2011-05-15 00:22:20

Re: Permutaions and combinations

#9 2011-05-15 00:38:04

Re: Permutaions and combinations

#10 2011-05-15 02:15:44

Re: Permutaions and combinations

#11 2011-05-15 02:30:26

Re: Permutaions and combinations

#12 2011-05-15 02:32:23

Re: Permutaions and combinations

#13 2011-05-15 03:22:21

Re: Permutaions and combinations

#14 2011-05-15 04:11:10

Re: Permutaions and combinations

#15 2011-05-15 07:32:47

Re: Permutaions and combinations

#16 2011-05-15 17:41:39

Re: Permutaions and combinations

#17 2011-05-15 19:17:55

Re: Permutaions and combinations

#18 2011-05-15 19:58:17

Re: Permutaions and combinations

#19 2011-05-15 20:02:56

Re: Permutaions and combinations

#20 2011-05-15 20:05:04

Re: Permutaions and combinations

#21 2011-05-15 20:35:06

Re: Permutaions and combinations

#22 2011-05-15 20:52:05

Re: Permutaions and combinations

#23 2011-05-15 22:47:13

Re: Permutaions and combinations

#24 2011-05-15 22:56:55

Re: Permutaions and combinations

#25 2011-05-25 20:16:42

Re: Permutaions and combinations

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