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Re: There are n identical red balls & m identical green balls. The number [#permalink]
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Yes it's the Pascal Rule. It is used in one of the official question too.

(n+m+1)Cm represents selecting the m objects out of n+m+1 directly.

We can do it another way too. Split n+m+1 in n+m objects and 1 object. In one case, we select m objects from n+m objects and in other case we select m-1 objects from n+m objects and select that 1 object we spited earlier. Total ways= (n+m)C(m-1)+(n+m)Cm

We can prove it algebraically too


(n+m)C(m-1)+(n+m)Cm

= \(\frac{(n+m)!}{(m-1)!(n+1)!} \)+ \(\frac{(n+m)!}{(m)!(n)!}\)

\(= \frac{(n+m)!}{(m-1)!n!} [\frac{1}{n+1} + \frac{1}{m}]\)

=\( \frac{(n+m)!}{(m-1)!n!} [\frac{m+n+1}{(n+1)m}]\)

\(= \frac{(n+m+1)!}{m!(n+1)!}\)

= n+m+1Cm


rsrighosh wrote:
nick1816

How did you reach from

(n+m)C(m-1)+(n+m)Cm to (n+m+1)Cm

I saw that in previous step as well. Is there any such rule?
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Re: There are n identical red balls & m identical green balls. The number [#permalink]
Expert Reply
Bunuel wrote:
There are n identical red balls & m identical green balls. The number of different linear arrangements consisting of “n red ball but not necessarily all the green balls’ is xCy then

(A) x = m + n, y = m
(B) x = m + n + 1, y = ,m
(C) x = m + n + 1, y = m + 1
(D) x = m + n, y = n
(E) x = m + n, y = n + 1




Solution:

If the linear arrangement consists of n red balls and no green balls, then there are n! / n! = nCn ways.

If the linear arrangement consists of n red balls and 1 green ball, then there are (n + 1)! / n! = (n+1)Cn arrangements.

If the linear arrangement consists of n red balls and 2 green balls, then there are (n + 2)! / (n! x 2!) = (n+2)Cn arrangements.



If the linear arrangement consists of n red balls and m - 1 green balls, then there are
(n + m - 1)! / (n! x (m - 1)!) = (n+m-1)Cn arrangements.

If the linear arrangement consists of n red balls and m green balls, then there are
(n + m)! / (n! x m!) = (n+m)Cn arrangements.

Now, we need to add up these numbers of arrangements.

nCn + (n+1)Cn + (n+2)Cn + … + (n+m-1)Cn + (n+m)Cn

Of course, we need to simplify this expression. However, if we don’t know how to do that algebraically, we can always do it arithmetically. That is, let’s use easy numbers for n and m. So, let’s say n = 2 and m = 4, we have:

2C2 + 3C2 + 4C2 + 5C2 + 6C2 = 1 + 3 + 6 + 10 + 15 = 35

Now, let’s see which answer choice (or choices) will yield 35 when n = 2 and m = 4:

A) x = 4 + 2 = 6, y = 4 → xCy = 6C4 = 15 → This is not 35.

B) x = 4 + 2 + 1 = 7, y = 4 → xCy = 7C4 = 35 → This IS 35.

C) x = 4 + 2 + 1 = 7, y = 4 + 1 = 5 → xCy = 7C5 = 21 → This is not 35.

D) x = 4 + 2 = 6, y = 2 → xCy = 6C2 = 15 → This is not 35.

E) x = 4 + 2 = 6, y = 2 + 1 = 3 → xCy = 6C3 = 20 → This is not 35.

Since the only expression that produces 35 when n = 2 and m = 4 is the expression in answer choice B, it is the correct answer.

Answer: B
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Re: There are n identical red balls & m identical green balls. The number [#permalink]
nick1816 wrote:
Number of arrangements of n red balls = 1=nC0
Number of arrangements of n red balls and 1 green ball= (n+1)C1
.......
Number of arrangements of n red balls and m green ball= (n+m)Cm

Total possible arrangements

= nC0+(n+1)C1+(n+2)C2+.......+(n+m)Cm

= (n+1)C0+(n+1)C1+ (n+2)C2+.......+(n+m)Cm

= (n+2)C1+(n+2)C2+.....+(n+m)Cm

Pattern goes on till

=(n+m)C(m-1)+(n+m)Cm

=(n+m+1)Cm

 

­Hi, 
Here in your solution, can you please explain how did you get (n+1)C0 from nC0?

Total possible arrangements
= nC0+(n+1)C1+(n+2)C2+.......+(n+m)Cm

= (n+1)C0+(n+1)C1+ (n+2)C2+.......+(n+m)Cm

= (n+2)C1+(n+2)C2+.....+(n+m)Cm

Thank you.
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Re: There are n identical red balls & m identical green balls. The number [#permalink]
Expert Reply
nC0 means ways to choose 0 out of n things. It will be just 1 way.
There fore be it nC0 or (n+1)C0 or 100C0, the value is same. But (n+1)C0 helps in simplifying the expression through Pascals law. nC1+nC2=(n+1)C2=>…. 10C1+10C2 = 11C2
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Re: There are n identical red balls & m identical green balls. The number [#permalink]
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