An Even Challenge

Probability Level 2

How many numbers of Six digits can be made by arranging the digits of the number $123425$ so that all the even digits do not occupy consecutive places.

Details and Assumptions

The number is indeed $123425$ , there is no typo.

Solve more problems like this in my set Combinatorics .

720 288 360 216

1 solution

Nikhil Ramakrishnan
Feb 28, 2015

The total number of possible six digit numbers that can be formed using the digits 1,2,2,3,4,5 is given by $\frac{6!}{2!}$ $=360$ Now, to find out the number of cases in which all the even digits are consecutive, we consider the digits 2,2 and 4 as one group. The total number of elements now is 4. $\boxed{224}135$

The possible cases is given by $\frac{4! \times 3!}{2!}$ $=72$

Therefore, the final result $360-72 = 288$

To reduce the number of arrangements in both cases, why is it divided by 2! ?

Ajit Deshpande - 6 years, 3 months ago

Because the number "2" occurs twice and its arrangement has to be excluded (for example, if arrangements of the digits 1,2,2 are considered, 212 will occur twice). Therefore we divide by 2! in both cases.

Nikhil Ramakrishnan - 6 years, 3 months ago

An Even Challenge

Solve more problems like this in my set Combinatorics .

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