q30

S

Saumyojit

Senior Member
Joined
Jan 21, 2020
Messages
1,032
How can I find the number of ways of selecting n articles out of 3n+1, out of which n are identical?

if n=2 , 3n+1=7 articles

A1 , A2, C, D , E , F, G (a1 a2 identical)

We select two at a time .
3n+1 c n gives us

Valid: A1A2 , A1C, A1d, A1e, A1f = 16

Invalid: A2C , A2D , A2E= 5

7c2=21 but we need to get 15 which is seven less than 21 .



if n=1 , 3n+1=4 articles
A1,A2,B,C
3n+1 c n gives us

Valid: A1, B, C
Invalid : A2

4C1: 4 . We need one less




What will be generalized Expression
 
Last edited:
To make sure we know exactly what the problem is, please show the entire original question, including any choices provided and the claimed answer if known, and tell us where it is from. It seems rather odd.

As I understand it, there are n identical items and 2n+1 distinct items (distinct both from one another and from the n), and you want to count the number of ways to choose n items (distinct or not). Is this correct?

So if n=1, all the items are distinct (contrary to what you show), and you have something like ABCD. There are 4 ways to choose 1.

If n=2, you have 2 identical items and 5 others, like AABCDEF. The possible choices of 2 are AA (1 way), A and any of the 5 (5 ways), and any pair of the 5 (10 ways), for a total of 16. Where do you get your 15? Are you implying that the n you choose must be distinct?
 
My friend told me Lets divide into two : identical and non identical group in to
n and 2n +1 articles respectively .

Identical n articles | Non identical 2n+1 articles
selection:
0 | 2n+1 c N
1 | 2n+1 c n-1
2 | 2n+1 C n-2
.. .
...
n | 2n+1 C 0

Each different event : adding up : 2n +1 c N + 2n +1 c n-1 + ......+ 2n+1 C 0

THen what ?
 
Saumyojit said:
My friend told me Lets divide into two : identical and non identical group in to
n and 2n +1 articles respectively .

Identical n articles | Non identical 2n+1 articles
selection:
0 | 2n+1 c N
1 | 2n+1 c n-1
2 | 2n+1 C n-2
.. .
...
n | 2n+1 C 0

Each different event : adding up : 2n +1 c N + 2n +1 c n-1 + ......+ 2n+1 C 0

THen what ?
Click to expand...
First, I'll tell you that I haven't even tried to fully solve this yet myself, because I'm here just to help you think, and don't want to have a solution in mind to short-circuit that process. But I have tried thinking along with you, and there are a couple things I would do if I were you. (In fact, I've just now done these bits.)

One is this: we've manually counted the answers for a couple small cases, so you can see whether any of the choices you are given yield those numbers. I normally despise multiple-choice problems, because they allow you to solve them without doing the real work; but one reason I ask for the choices is that they can help us determine whether our interpretation is right, and they can also give us a sense of how complicated an answer we can expect. In this case, with "none of these" as an option, we can't be sure, but it can still help a little. In any case, we already know which answer it has to be, unless they are being tricky and that answer only works for the first two cases!

Another is, now that you have a proposed general solution, though it is not simple enough to check against the choices, we can check it against our answers! Does your summation give 4 for n=1 and 16 for n=2?

At this point, we've been encouraged that the summation looks like it is correct, and should be able to be simplified; so we have two choices: We can work algebraically (or otherwise) to simplify it, or we can look for a different approach that won't lead to a summation (perhaps using the apparent answer as a clue to what sort of approach might lead to that sort of answer!). Or we could suppose that the apparent answer (suggested not only by the choices but also by our small cases) is correct, and use mathematical induction to prove that the summation is equal to that, if you know that method. (It's a method that is used to prove formulas but can't be used to derive them.)

So, do what I've suggested (checking the choices and checking your summation), and then decide what to do next! (I'll probably be doing that now. I have several methods in mind.)
 
sky2rain said:
Imagine you have a passcode lock with 5 dials: ABCDE, on first dial you have 0-5, second 0-4… and last just 0 and 1. Random variable requires permutations that add up to 5 and how many cases are there.
Click to expand...
I think you put this in the wrong thread. Did you intend it for "q31" (which is about the first question in the image in #3)?

That's not to say that your answer is right.
 
Saumyojit said:
I dont know Induction .

Answer is 2 ^ 2n

n=2 then it gives 16 . So 2n +1 c N + 2n +1 c n-1 + ......+ 2n+1 C 0 have to give 2^2n
Click to expand...
The easiest way to see that {2n +1} C n + {2n +1} C {n-1} + ... + {2n+1} C 0 = 2^{2n} is to think in terms of Pascal's triangle. The sum is exactly half a row, and the sum of a row is a power of 2. So the sum is 2^{2n+1} / 2 = 2^{2n}.

The sum of a row can be seen by realizing that it is the total number of ways to make any subset (of any size), which in this case is 2^{2n+1}. This can probably be turned into a different way to look at the original problem, but it doesn't seem obvious to me. I don't know what approach the authors might expect you to use.

By the way, please try to use parentheses at least occasionally!
 
Saumyojit said:
i know pascal triangle --> 1 , 11 ,121 , 1331 etc
But i did not understand the connection?
Click to expand...
What do you understand about it?

I told you that my approach depends on knowing a specific fact, namely that 1+1 = 2^1, 1+2+1 = 2^2, 1+3+3+1 = 2^3, and so on.

Do you see that your sums are 1 and 1+3, halves of 1+1 and 1+3+3+1? And have you thought about the explanation I gave?
 
Dr.Peterson said:
The easiest way to see that {2n +1} C n + {2n +1} C {n-1} + ... + {2n+1} C 0 = 2^{2n} is to think in terms of Pascal's triangle. The sum is exactly half a row, and the sum of a row is a power of 2. So the sum is 2^{2n+1} / 2 = 2^{2n}.
Click to expand...
Now i understood .

{2n +1} C n + {2n +1} C {n-1} + ... + {2n+1} C 0 = if n=2 , 2n+1=5 then we are considering only half of the power set of 2n+1 .

THANKS
 
I have been reluctant to post this because you may not know generating polynomials.
But look at this in the expansion of [imath](1+t)(1+t+t^2)(1+t+t^2+t^3)(1+t+t^2+t^3+t^4)(1+t+t^2+t^3+t^4+t^5)[/imath] each term give us a count the term [imath]71 t^5[/imath] tell us that there are seventy-one ways to pick five using the given restrictions.
 
pka said:
I have been reluctant to post this because you may not know generating polynomials.
But look at this in the expansion of [imath](1+t)(1+t+t^2)(1+t+t^2+t^3)(1+t+t^2+t^3+t^4)(1+t+t^2+t^3+t^4+t^5)[/imath] each term give us a count the term [imath]71 t^5[/imath] tell us that there are seventy-one ways to pick five using the given restrictions.
Click to expand...
This is an answer to q31:

q31

In how many ways can a selection be made of 5 letters out of 5As ,4 Bs, 3Cs, 2 Ds and 1E? I could not find any approach
www.freemathhelp.com

Both questions are shown in #3 here, which is causing confusion! (See #6 here.)
 
You must log in or register to reply here.
Top