q29

[Saumyojit]

The number of ways in which mn students can be distributed equally among n sections is-

If m is 2 and n =3 mn =6​

THen , it would be like (6c2 * 4c2 * 2c2 )/6 = 15

(mnCm * mn-mCm* mn-m - m Cm*....mcm)/ n!

So it can be mn C m as 6c2=15

 

[pka]

The number of ways in which mn students can be distributed equally among n sections is-​

​

If m is 2 and n =3 mn =6​

THen , it would be like (6c2 * 4c2 * 2c2 )/6 = 15

(mnCm * mn-mCm* mn-m - m Cm*....mcm)/ n!

So it can be mn C m as 6c2=15

The six students can appear on a roster in alphabetical order.
Let [imath]A~\&~B[/imath] be the two sections of the course.
How many ways can the string [imath]a\,a\,a\,b\,b\,b[/imath] be rearranged?
We have already helped you with this concept.

 

[Saumyojit]

How many ways can the string a a a b b b be rearranged?

6!/ (3!*3!) = (mn)! / (m! * m!) Is it correct?

 

[Saumyojit]

Actually this is the question . acc to my logic (d) is correct as i have stated with mn=6 eg above .

 

[Dr.Peterson]

6!/ (3!*3!) = (mn)! / (m! * m!) Is it correct?

That's a hasty generalization! Your example case is rather simple, with only two groups.

View attachment 28999


Actually this is the question . acc to my logic (d) is correct as i have stated with mn=6 eg above .

How does that agree with your example, mn=6?

(I can hardly read the small print -- please make a better image.)

Try another example, say m=3, n=4. Or, try thinking through the general case carefully, as with the specific case. And try explaining it fully to us, so you can see whether everything makes sense.

Never leap from one thought to another without thinking.

 

[Saumyojit]

Your example case is rather simple, with only two groups.

three groups of two each.

say m=3, n=4.

yes i did . Both are not matching you are right but if m=3, n=4
Then this expression is right? --> (12c3 * 9c3 * 6c3 * 3c3)/ 4! . I am dividing by 4 factorial as all 4 groups are same so these two combo are the same

ABC DEF GHI JKL and ABC GHI JKL DEF etc .

What will be the answer ?

 

[Dr.Peterson]

I am dividing by 4 factorial as all 4 groups are same so these two combo are the same

ABC DEF GHI JKL and ABC GHI JKL DEF etc .

Interesting thought; but I would take the sections (groups) as being distinguishable, as they would be in a class, so that those would be the same. We'll have to see at the end whether any of the choices agree with either interpretation.

Then this expression is right? --> (12c3 * 9c3 * 6c3 * 3c3)/ 4! .

I don't know. Tell me what each part means. (As you should know, there can be many different ways to write equivalent answers.)

pka's model of the problem is a very good approach to take, and will probably be easier to generalize than what you are doing. Have you tried applying it?

What will be the answer ?

That's your job. I'm just here to nudge you back on track.

 

[pka]

6!/ (3!*3!) = (mn)! / (m! * m!) Is it correct?

Why is it that you want a quick answer at the expense of not having the foggiest notion of the general idea.
Example; Suppose we have one hundred students to be placed into five sections of a course.
Each section has twenty students. Name the sections [imath]A\,B\,C\,D\,E[/imath].
If we have a roster of names in alphabetical order. And we need to assign the students to the sections randomly.
Using twenty each of the letters [imath]A\,B\,C\,D\,E[/imath] we can make a string of one hundred.
How many possible such strings are there? Well [imath]\dfrac{(100)!}{(20!)^5}[/imath].

But in your post #4, the printed question says: [imath]\text{In how many ways can }\bf{m\,n}\text{ things be distributed equally among }n\,\text{ groups ?}[/imath].
Now that is vague at best. Are the things identical such as steel ball-barings; or distinguishable like the students. Are the groups distinguishable like the sections of a course or they like studygroups where only the student content makes a difference. Each of these types has a different way of being counted. If we are not told these thing in the problem then there is no way to answer the question. So tell your lecturer to add the essential details needed to answer the question.

 

[Saumyojit]

Using twenty each of the letters A B C D EA\,B\,C\,D\,EABCDE we can make a string of one hundred.
How many possible such strings are there? Well (100)!(20!)5\dfrac{(100)!}{(20!)^5}(20!)5(100)!.

absolutely correct. (m*n)! / (m!)^n
The sections are Distinguishable and so are students .

Are the things identical such as steel ball-barings; or distinguishable like the students.

As they have not mentioned so we assume indistinguishable.

Are the groups distinguishable like the sections of a course or they like studygroups where only the student content makes a difference.

I assume not as they are Groups of equal qunatity not SECTIONS.

don't know. Tell me what each part means.

If m is 2 and n =3 mn =6​

THen , it would be like (6c2 * 4c2 * 2c2 )/6 = 15

Selecting any two things out of six and putting in one group and then two out of 4 putting in second group , then two out of two in last group .

Dividing by 3! i have already gave you reason .

Now tell me what will be the approach of image Quest.

 

[Saumyojit]

Now it makes sense

(mnCm * mn-mCm* mn-m - m Cm*....mcm)/ n!
ok this is equivalent to writing (mn)! / (m!* n!)
Groups are same so divided by 3! and rearrangmenet in between is not needed thats why 2!

 

[Dr.Peterson]

The sections are Distinguishable and so are students .

I assume not as they are Groups of equal qunatity not SECTIONS.

Now I'm very confused. I understood the questions in #1 and #4 to be your attempts at asking the same question, since you put them in the same thread implying that your previous work applied to both problems. You need to stop doing this sort of thing. State the actual question from the start.

But I agree with pka, that the version in #4 doesn't tell us what is or is not distinguishable, so it is a poorly worded question.

Now tell me what will be the approach of image Quest.

Who is Image Quest, and why would any of us know what they would do??