Consider the $$10$$-digit number $$M=9876543210$$. We obtain a new $$10$$-digit number from $$M$$ according to the following rule: we can choose one or more disjoint pairs of adjacent digits in $$M$$ and interchange the digits in these chosen pairs, keeping the remaining digits in their own places. For example, from $$M=9876543210$$, by interchanging the $$2$$ underlined pairs, and keeping the others in their places, we get $$M_{1}=9786453210$$. Note that any number of (disjoint) pairs can be interchanged. Find the number of new numbers that can be so obtained from $$M$$.

Number the places in $$M=9876543210$$ from $$1$$ (left-most) to $$10$$ (right-most).

We may choose any collection of adjacent pairs $$(1,2),\,(3,4),\ldots,(9,10)$$ provided no two chosen pairs overlap. Each chosen pair is then swapped once. Because all digits of $$M$$ are different, every distinct set of chosen pairs produces a unique final 10-digit string, and every such string arises from exactly one set of pairs. Hence the problem reduces to counting the possible selections of disjoint adjacent pairs in a row of $$10$$ positions.

Let $$f(n)$$ denote the number of ways to select disjoint adjacent pairs among $$n$$ consecutive positions.

Initial values:
$$f(0)=1$$  (no position, one empty selection)
$$f(1)=1$$  (a single position cannot be paired)

For $$n\ge 2$$ consider position $$1$$:

Thus we obtain the Fibonacci recurrence
$$f(n)=f(n-1)+f(n-2)\quad (n\ge 2).$$

Building the sequence up to $$n=10$$:
$$\begin{aligned} f(2)&=f(1)+f(0)=1+1=2,\\ f(3)&=f(2)+f(1)=2+1=3,\\ f(4)&=5,\; f(5)=8,\; f(6)=13,\; f(7)=21,\; f(8)=34,\; f(9)=55,\; f(10)=89. \end{aligned}$$

The value $$f(10)=89$$ counts even the “do nothing” choice where no pair is selected; that choice leaves the number unchanged as $$9876543210$$, which is not a new number.

Therefore the number of new 10-digit numbers obtainable is
$$89-1=88.$$

Final answer: 88