There are n identical red balls & m identical green balls. The number

Question

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

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nick1816 · Answer

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

rsrighosh · Answer

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?

nick1816 · Answer

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{(n+m)!}{(m-1)!n!}

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

= n+m+1Cm

ScottTargetTestPrep · Answer

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

zlishz · Answer

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.

chetan2u · Answer

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

Kinshook · Answer

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

The number of arrangements = nC0 + (n+1)C1 + (n+2)C2 + (n+3)C3 + .... (n+m)Cm =

In general (n+x)Cx + (n+x)C(x+1) = (n+x)!/x!n! + (n+x)!/(x+1)!(n-1)! = (n+x)!/x!(n-1)! (1/n + 1/x+1) = (n+x)!/x!(n-1)! (n+x+1)/n(x+1) = (n+x+1)!/n!(x+1)! = (n+x+1)C(x+1)

The number of arrangements = (n+2)C1 + (n+2)C2 + (n+3)C3 + .... (n+m)Cm = (n+3)C2 + (n+3)C3 + .... (n+m)Cm = (n+4)C3 + ...(n+m)Cm = (n+m+1)Cm

IMO B

There are n identical red balls & m identical green balls. The number

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There are n identical red balls & m identical green balls. The number

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