The number of 5-digit numbers in which no two consecutive digits are identical is
A.
$${9^2} \times {8^3}$$
B.
$$9 \times {8^4}$$
C.
$${9^5}$$
D.
None of these
Answer :
$${9^5}$$
Solution :
Ways: \[\begin{array}{*{20}{c}}
\times \\
9
\end{array}\,\,\begin{array}{*{20}{c}}
\times \\
9
\end{array}\,\,\begin{array}{*{20}{c}}
\times \\
9
\end{array}\,\,\begin{array}{*{20}{c}}
\times \\
9
\end{array}\,\,\begin{array}{*{20}{c}}
\times \\
9
\end{array}\]
0 cannot be placed in the first place. In the next place any digit except the one used in the first place can be used, e.t.c.
Releted MCQ Question on Algebra >> Permutation and Combination
Releted Question 1
$$^n{C_{r - 1}} = 36,{\,^n}{C_r} = 84$$ and $$^n{C_{r + 1}} = 126,$$ then $$r$$ is:
Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated are
Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First the women choose the chairs from amongst the chairs marked 1 to 4 ; and then the men select the chairs from amongst the remaining. The number of possible arrangements is