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  1. In how many ways can a given positive integer $n \geq 2$ be expressed as the sum of $2$ positive integers (which are not necessarily distinct). For example, for $n=3$, the number of ways is $2$, i.e., $1+2, 2+1$. Give only the answer without any explanation.
  2. In how many ways can a given positive integer $n \geq 3$ be expressed as the sum of $3$ positive integers (which are not necessarily distinct). For example, for $n=4$, the number of ways is $3$, i.e., $1+2+1, 2+1+1$ and $1+1+2$. Give only the answer without explanation.
  3. In how many ways can a given positive integer $n \geq k$ be expressed as the sum of $k$ positive integers (which are not necessarily distinct). Give only the answer without explanation.
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54 votes
  1. $n= 2 \left(1+1\right),\;n=3\left(1+2, 2+1\right),\\n=4\left(1+3,3+1,2+2\right),\;n=5\left(1+4,4+1,2+3,3+2\right)$
    so $x_1+x_2=n\;\text{and}\;x_1,x_2>{0}$ (no.of integral sol)
    This is same as number of ways of putting $\left(n-2\right)$  (as we can't have $0$ for either $x_1$ or $x_2$) identical balls into two distinct bins, which is obtained by putting a divider across $\left(n-2\right)$ balls and taking all possible permutations with $\left(n-2\right)$ being identical. i.e., $\frac{(n-2 + 1)!}{(n-2)!} = (n-1).$
    We can also use the following formula ,
    $^{(n-2+2-1)}C_{(2-1)}=^{n-1}C_1.$
     
  2. $n=3\left(1+1+1\right),\;n=4\left(1+1+2,1+2+1,2+1+1\right),\\ n=5\left(1+1+3,1+3+1,3+1+1,2+2+1,2+1+2,1+2+2\right) $
    so $x_1+x_2+x_3=n\;\text{and}\;x_1,x_2,x_3>0$ (no.of integral sol) 
    Here, we can permute $\left(n-3\right)$ items with $2$ dividers which will give $\frac{(n-3 + 2)!}{(n-3)!2!}$
    $\begin{align}&=\frac{\left(n-1\right)!}{\left(n-1-2\right)!2!}\\\\&=\;^{n-1}C_2\end{align}$
     
  3. $^{(n-k+k-1)}C_{k-1}=^{n-1}C_{k-1}.$
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8 votes
8 votes

We know that the no. of +ve integral solution to the equation $x_{1} + x_{2} + x_{3}+ ... + x_{k} = n$ is given by $\binom{n-1}{k-1}$.

(a) Let, $x_{1}$ and $x_{2}$ be the two positive integers (not necessarily distinct) such that their sum is equal to $n\ (n≥2)$.

Therefore, $x_{1} + x_{2} = n$. 

Now, we have to find all such values of $x_{1}$ and $x_{2}$ that satisfy the above equation. In other words, the problem basically reduces to  finding the no. of +ve integral solution to the above equation.

Therefore, the req. no. of ways is $\binom{n-1}{2-1} = \binom{n-1}{1}$. 

(b) Let, $x_{1}$, $x_{2}$ and $x_{3}$ be the three positive integers (not necessarily distinct) such that their sum is equal to $n\ (n≥3)$.

Therefore, $x_{1} + x_{2} + x_{3} = n$. 

Now, we have to find all such values of $x_{1}$, $x_{2}$ and $x_{3}$ that satisfy the above equation. In other words, the problem basically reduces to  finding the no. of +ve integral solution to the above equation.

Therefore, the req. no. of ways is $\binom{n-1}{3-1} = \binom{n-1}{2}$.

(c) Let, $x_{1}$, $x_{2}$, $x_{3}$,...,$x_{k}$ be the $k$ positive integers (not necessarily distinct) such that their sum is equal to $n\ (n≥k)$.

Therefore, $x_{1} + x_{2} + x_{3}+ ... + x_{k} = n$. 

Now, we have to find all such values of $x_{1}$, $x_{2}$, $x_{3}$,...,$x_{k}$ that satisfy the above equation. In other words, the problem basically reduces to  finding the no. of +ve integral solution to the above equation.

Therefore, the req. no. of ways is $\binom{n-1}{k-1}$.

      

3 votes
3 votes
SIMPLY APPLY THE LOGIC:

x1+x2+x3+x4+...........xn=r   where x1,x2.....xn>=0    apply c(n+r-1,r)  

make similarly all the 3 cases in order that it satisfies the equation

a-  x1+x2=n where x1,x2>=1 so make it x1+x2=n-2 now x1,x2>=0 so c(n-2+2-1,1)=n-1

b-x1+x2+x3=n where x1,x2,x3>=1 so x1+x2+x3=n-3 so c(n-3+3-1,2)=c(n-1,2)

c-x1+x2+.....xk=n where x1,x2,x3....xk>=n so x1+x2+x3+....xk=n-k=c(n-k+k-1,k-1)=c(n-1,k-1)
1 votes
1 votes

(a) Total no of ways = C(n-1 ,1) = n-1

(b)Total no of ways = C(n-1 ,n-3) = C(n-1 ,2) = (n-1)⨉(n-2) /2

(c)Total no of ways = C(n-k+k-1 ,n-k) = C(n-1 , n-k) = C(n-1 , k-1)

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