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\[{}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}\]

\[N = \]the total number of elements in a set

\[K = \] The number of selected objects (the order of the objects is not important)

\[! = factorial\]

The variable name consists only of a letter, then that letter can be chosen in 26 ways from the alphabets.

If the code is created using a letter followed by a digit, then code can be created in

\[{}^{26}{C_1} \times {}^{10}{C_1}\]

These two events are independent

Now,

Above combination can be done in only one way.

\[1! = 1\]

\[ \Rightarrow \dfrac{{26!}}{{1!(26 - 1)!}} \times \dfrac{{10!}}{{1!(10 - 1)!}} \times 1\] \[\left[ {{}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}} \right]\]

\[ \Rightarrow \dfrac{{26.25!}}{{25!}} \times \dfrac{{10.9!}}{{9!}}\] \[\left[ \begin{gathered}

n = 26 \\

r = 1 \\

\end{gathered} \right]\]

\[ \Rightarrow 26 \times 10\]

\[ \Rightarrow 260\]

Hence

Total number of different variable names that can exist in that language is

\[ = 26 + 260\]

\[ = 286\]

\[10! = 1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9 \times 10\]

So, it can be writing as

\[10! = 9!\, \times 10\]

1)(Remember)\[1! = 1\]

2) In the above question, it is fixed followed by a decimal digit, So the rotation can be done by only 1 way.