<s>[INST] $5$ digit no. in which at least $3$ digits are identical. The number of $5$-digit numbers that can be made with digits $\left\{1,2,3,4,5,6\right\}$
in which at least $3$ digits are identical is.
$\bf{My\; Try::}$ 
No.,s in which at least $3$ digits are Identical is $ = $ Total -no digit are identical - exactly $2$ digit are 
identical.
So Total no. of ways is $ = 6\times 6 \times 6 \times 6 \times 6=6^5$ 
and no digit are identical is $ = 6\times 5 \times 4 \times 3 \times 2 = 6!$
and I did not understand how can i find the cases in which exactly $2$ digits are identical.
Help me
Thanks
 [/INST] There are 3 cases:
1. 3 digits identical: $aaabc$
There are $4+3+2+1$ ways of arranging $b$ and $c$ such that $b$ always comes before $c$. Then there are $6$ choices for $b$, $5$ for $c$ and $4$ for $a$. Thus there are $10\cdot6\cdot5\cdot4=1200$ numbers.
For $aaabb$, we have $10$ ways of arrangement and therefore a total of $10\cdot6\cdot5=300$ numbers.
2. 4 digits are identical: $aaaab$
There are $5$ positions for $b$. So there $5\cdot6\cdot5=150$ numbers.
3. 5 digits are identical: $aaaaa$
There are clearly $6$ such numbers.
So there are $\boxed{1656}$ such numbers.

To count the ones with $2$ identical digits, we have $2$ cases.
1. $aabcd$
There are ${5\choose 2}$ ways of placing the $a$'s. We will not arrange $bcd$ since it will cause over counting. So we have $10\cdot6\cdot5\cdot4\cdot3=3600$ such numbers.
2. $aabbc$
There are $5$ ways for placing the $c$. For the remaining $4$ positions we have $4\choose 2$ ways of placing the $a$'s and $1$ way of placing the $b$'s. So there are $5\cdot 6$ ways of arranging. But we divide this by $2$ since $a$ and $b$ are interchangeable. So we have $15$ ways of arranging these. Thus there are $15\cdot6\cdot5\cdot4=1800$ numbers.
So there are $5400$ numbers with $2$ digits identical. 
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