Question

let N=abcdefghij be a10 digit natural nuber uch that a<=b<=c<=d<=e<=f<=g<=h<=i>=j let m be number of such numbers such that all digits belong to {1,2,3,4,5} let n be number of such numbes formued usong all the diits 1,2,3,4,5. m+n?

Answer 1

3863

Answer 2

We are given a 10-digit natural number

$N = a\,b\,c\,d\,e\,f\,g\,h\,i\,j$

with the restrictions

$a\le b\le c\le d\le e\le f\le g\le h\le i$ and $i\ge j$.

All digits are chosen from the set $\{1,2,3,4,5\}$.

Define:

• $m$ = number of such numbers (with digits from $\{1,2,3,4,5\}$) satisfying the above order-conditions.
• $n$ = number of such numbers (again with the above order-conditions) that use all the digits $1,2,3,4,5$ at least once.

We want to compute $m+n$.

Step 1. Counting $m$

A. Count the first 9 digits

Since the first 9 digits are non-decreasing and come from 5 digits, the total number of ways is given by the stars and bars formula:

$\text{Total}_{9} = \binom{5+9-1}{9} = \binom{13}{9} = \binom{13}{4} = 715$.

However, the final digit $j$ is not free; it must satisfy $j\le i$ where $i$ (the 9th digit) depends on the chosen 9-tuple.

B. Count by conditioning on the 9th digit

For each non-decreasing 9-tuple the 9th digit (say $k$, with $k\in\{1,2,3,4,5\}$) is fixed. Then $j$ can be any digit in $\{1,2,\dots,k\}$ (there are $k$ choices).

Now, we need to sum over the cases when the last digit (of the 9-tuple) is exactly $k$.

Let $f(k)$ = number of non-decreasing sequences of length 9 from $\{1,2,3,4,5\}$ with 9th digit exactly $k$.

One way to count $f(k)$ is to note:

$f(k)= \underbrace{\binom{9+k-1}{9}}_{\text{all sequences using }1,2,\dots,k} - \underbrace{\binom{9+k-2}{9}}_{\text{sequences using }1,2,\dots,k-1}$.

For each $k$, the total contribution is $k\cdot f(k)$. Hence,

$m = \sum_{k=1}^{5} k\left[\binom{9+k-1}{9} - \binom{9+k-2}{9}\right]$.

Let’s calculate term-by-term:

• For $k=1$:

  $f(1)= \binom{9}{9}-\binom{8}{9} =1-0=1$, contribution= $1\cdot1=1$.

• For $k=2$:

  $f(2)= \binom{10}{9}-\binom{9}{9}=10-1=9$, contribution= $2\cdot9=18$.

• For $k=3$:

  $f(3)= \binom{11}{9}-\binom{10}{9} = \binom{11}{2}-\binom{10}{1}=55-10=45$, contribution= $3\cdot45=135$.

• For $k=4$:

  $f(4)= \binom{12}{9}-\binom{11}{9} = \binom{12}{3}-\binom{11}{2}=220-55=165$, contribution= $4\cdot165=660$.

• For $k=5$:

  $f(5)= \binom{13}{9}-\binom{12}{9} = \binom{13}{4}-\binom{12}{3}=715-220=495$, contribution= $5\cdot495=2475$.

Summing all contributions:

$m = 1+18+135+660+2475 = 3289$.

Step 2. Counting $n$

Here we count those sequences from $m$ that use all the digits $\{1,2,3,4,5\}$ at least once.

Let $A_d$ be the set of sequences (satisfying our order conditions) that do not contain digit $d$. Using inclusion-exclusion, if we define for a set $S\subseteq \{1,2,3,4,5\}$ of missing digits, note that when the allowed set has $r$ digits, the count (in place of $m$) is, by a similar argument, a function $f(r)$ with:

• $f(5)=3289$ (we computed this).

• $f(4)$: This is the count when the digits come from any 4-element subset. Repeating the above process (or by analogy) we obtain:

  $f(4)= 1+18+135+660 = 814$.

• $f(3)= 1+18+135 = 154$.

• $f(2)= 1+18 = 19$.

• $f(1)= 1$.

Now by inclusion-exclusion,

$n = f(5) - \binom{5}{1}f(4) + \binom{5}{2}f(3) - \binom{5}{3}f(2) + \binom{5}{4}f(1)$,

since $f(0)=0$.

Plugging in:

$n = 3289 -5\cdot814 +10\cdot154 -10\cdot19 +5\cdot1$.

Calculate:

$5\cdot814 = 4070$, $10\cdot154 = 1540$, $10\cdot19=190$, $5\cdot1=5$.

Thus,

$n = 3289 -4070 +1540 -190 +5 = (3289 -4070) +1540 -190 +5 = -781 +1540 -190 +5$.

$-781+1540 =759$, $759-190 =569$, $569+5=574$.

So, $n = 574$.

Step 3. Final Answer

$m+n = 3289 + 574 = 3863$.

Therefore, the final answer is 3863. The question type is integer and the difficulty is medium.