The number of 6-digit numbers that can be made with the digits 0,1, 2, 3, 4 and 5 so that even digits occupy odd places, is
A.
24
B.
36
C.
48
D.
None of these
Answer :
24
Solution :
$$ \times | \times | \times |$$ Crosses can be filled in $$^3{P_3} - {\,^2}{P_2}$$ ways ( $$\because 0$$ cannot go in the first place from the left).
The remaining places can be filled in 3! ways.
∴ the required number of numbers $$ = \left( {^3{P_3} - {\,^2}{P_2}} \right) \times 3!.$$
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