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Consider the set of all functions from $\{1, 2, . . . ,m\}$ to $\{1, 2, . . . , n\}$,where $n > m$. If a function is chosen from this set at random, the probability that it will be strictly increasing is

  1. $\binom{n}{m}/n^m\\$
  2. $\binom{n}{m}/m^n\\$
  3. $\binom{m+n-1}{m-1}/n^m\\$
  4. $\binom{m+n-1}{m}/m^n$
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2 Answers

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5 votes
Best answer
Total no. of functions is $n^m$. [For every value from ${1,2....,n}$ there are m possibilities].

We choose an m-element subset of n, number of such subsets $=\binom{n}{m}$

Since it is an increasing function, we can order the elements in only one way, i.e in ascending order.

Therefore , probability is $\frac{\binom{n}{m}}{n^m}$
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4 Comments

@Arkaprava

Here condition is $n> m$

Suppose, $2$ questions,

$1)$ If there are $5$ children, and $2$ orange want to divide among them, possible?No.

$2)$ If there are $3$ children and $5$ orange, and each children gets $2$ orange possible?No.

So, how non-decreasing with repetitive order possible everytime?

0
0
n>m, but we are selecting m items from n types of items. Sort the elements in order, then you'll get your sequence.
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0
I havenot got. Plz solve those two questions:(
0
0
0 votes
0 votes
Surjective function (ONTO)

SO , nCm/ n^m

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