Question

How many six-digit numbers are there in which no digit is repeated, even digits appear at even places, odd digits appear at odd places and the number is divisible by 4 ?
(a) 3600
(b) 2700
(c) 2160
(d) 1440

Answer 1

Hint: There are five even digit numbers and five odd digit numbers. And also the number is divisible by 4 so in sixth place the number can be either 2 or 6. We will use the combination formula \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}\] to find the answer.

Complete step-by-step answer:
Before proceeding with the question, we should understand the definition of combination.
The combination is a way of selecting items from a collection, such that (unlike permutations) the order of selection does not matter. In smaller cases, it is possible to count the number of combinations. Combination refers to the combination of n things taken k at a time without repetition. So the formula for combination is \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!(n-r)!}........(1)\].

ODD NUMBER(FIRST PLACE)

EVEN NUMBER(SECOND PLACE)

ODD NUMBER(THIRD PLACE)

EVEN NUMBER(FOURTH PLACE)

ODD NUMBER(FIFTH PLACE)

EVEN NUMBER(SIXTH PLACE)

We need to fill the above spaces in such a way that second, fourth and sixth places have even numbers and the rest places have odd numbers. Also it is mentioned in the question that none of the numbers are repeated. Since the number is divisible by 4, the possible numbers for sixth place can be either 2 or 6. In fifth place there can be any one of the 5 odd numbers (1,3,5,7,9). In fourth place there can be one of the four even numbers. And in third place also there can be one of the 4 odd numbers left. Similarly, in second place there can be any one of the 3 even numbers left. Finally, in first place we can have any one of the 3 odd numbers left.
So the total number of six digit numbers \[={}^{3}{{C}_{1}}+{}^{3}{{C}_{1}}+{}^{4}{{C}_{1}}+{}^{4}{{C}_{1}}+{}^{5}{{C}_{1}}+{}^{2}{{C}_{1}}.......(2)\]
Now applying combination formula from equation (1) we get,
\[\begin{align}
  & \Rightarrow \dfrac{3!}{1!\times 2!}\times \dfrac{3!}{1!\times 2!}\times \dfrac{4!}{1!\times 3!}\times \dfrac{4!}{1!\times 3!}\times \dfrac{5!}{1!\times 4!}\times \dfrac{2!}{1!\times 1!} \\
 & \Rightarrow 3\times 3\times 4\times 4\times 5\times 2=1440 \\
\end{align}\]
Hence the number of six digit numbers is 1440. So the correct answer is option (d).
Note: We need to keep in mind that the digits are not being repeated so hence we are decreasing the numbers available to be filled after filling places consecutively. We may make a mistake in solving equation (2) so we need to remember the combination and factorial formula.