Permutations and Combinations 3 Question 11
11. Let $n _1<n _2<n _3<n _4<n _5$ be positive integers such that $n _1+n _2+n _3+n _4+n _5=20$. The number of such distinct arrangements $\left(n _1, n _2, n _3, n _4, n _5\right)$ is
(2014 Adv.)
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Answer:
Correct Answer: 11. (7)
Solution:
- PLAN Reducing the equation to a newer equation, where sum of variables is less. Thus, finding the number of arrangements becomes easier.
As, $\quad n _1 \geq 1, n _2 \geq 2, n _3 \geq 3, n _4 \geq 4, n _5 \geq 5$
Let $\quad n _1-1=x _1 \geq 0, n _2-2=x _2 \geq 0, \ldots, n _5-5=x _5 \geq 0$
$\Rightarrow$ New equation will be
$ \begin{aligned} & x _1+1+x _2+2+\ldots+x _5+5=20 \\ \Rightarrow \quad & x _1+x _2+x _3+x _4+x _5=20-15=5 \end{aligned} $
Now,
| $x _1 \leq x _2 \leq x _3 \leq x _4 \leq x _5$ | ||||
|---|---|---|---|---|
| $x _1$ | $x _2$ | $x _3$ | $x _4$ | $x _5$ |
| 0 | 0 | 0 | 0 | 5 |
| 0 | 0 | 0 | 1 | 4 |
| 0 | 0 | 0 | 2 | 3 |
| 0 | 0 | 1 | 1 | 3 |
| 0 | 0 | 1 | 2 | 2 |
| 0 | 1 | 1 | 1 | 2 |
| 1 | 1 | 1 | 1 | 1 |
So, 7 possible cases will be there.
