<s>[INST] Find the number of five digit numbers of the form $d_1d_2d_3d_4d_5$ and satisfying $d_1Find the number of five digit numbers of the form $d_1d_2d_3d_4d_5$ and satisfying $d_1<d_2\le d_3<d_4\le d_5.$

I considered $d_1,d_2,d_3,d_4,d_5$ as 5 baskets and these have to be filled with 0 to 9 apples.I found total ways as $\binom{10+5-1}{5-1}$ but the answer given is $\binom{11}{6}.$
 [/INST] Method 1:  The leading digit must be at least $1$, so $d_1, d_2, d_3, d_4, d_5 \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$. 
There are three possibilities:


*

*All the digits are different: There are $\binom{9}{5}$ such cases since selecting five of the nine digits completely determines the number since $d_1 < d_2 < d_3 < d_4 < d_5$

*Four different digits are used:  There are $\binom{9}{4}$ such cases in which $d_2 = d_3$ since the repeated digit must be the second smallest digit selected.  There are also $\binom{9}{4}$ such cases in which $d_4 = d_5$ since the repeated digit must be the largest digit selected.  Thus, there are $2\binom{9}{4}$ such cases.

*Three different digits are used:  There are $\binom{9}{3}$ such cases since the two repeated digits must be the two largest digits selected.


Total:  Since the three cases are disjoint, there are 
\begin{align*}
\binom{9}{5} + 2\binom{9}{4} + \binom{9}{3} & = \binom{9}{5} + \binom{9}{4} 
+ \binom{9}{4} + \binom{9}{3}\\
& = \binom{10}{5} + \binom{10}{4} && \text{by Pascal's Identity}\\
& = \binom{11}{5} && \text{by Pascal's Identity}
\end{align*}
numbers with the desired property.
Method 2:  Since the leading digit must be at least $1$, $d_1, d_2, d_3, d_4, d_5 \in \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$.
We can set up a bijection between selections of five numbers from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ satisfying $d_1 < d_2 \leq d_3 < d_4 \leq d_5$ and selections of five distinct elements $x_1, x_2, x_3, x_4, x_5$ of the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11\}$ by setting $x_1 = d_1$, $x_2 = d_2$, $x_3 = d_3 + 1$, $x_4 = d_4 + 1$, and $x_5 = d_5 + 2$.  For instance, the number $35579$ corresponds to the selection $\{3, 5, 6, 8, 11\}$.  Hence, the number of five-digit numbers with the desired property is $\binom{11}{5}$.
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