Permutations and Combinations - Online Test

You have two different kinds of such three-digit even numbers.First is  5 at the hundred'splace and  second 5 is not at the hundred'splace

•  In first case no is of the form 57x, where x is the unit's digit ,which can  be 0,2,4,6,8 which is just 5 possibilities.Hence the no of possibilities in this case is 1x1x5=5

• In second case the hundred's digit can be 1,2,3,,4,,6,7,8,or,9 which is 8 ways and the ten's digit  can be  any of the 9 numbers and unit digit can be any of the 5 even numbers .Therfore the no: of ways  will be 8×9×5=360

So total we  have 360 + 5 = 365 possibilities.

Total possibilities  of 5 digit numbers which can be formed using the given digits are 5!= 120 ways.

But since the number should be greater than 56000 we cannot have the numbers starting with 4 or 54

The combinations in which the number 4 comes at the start  is 4!  = 24 ways.

The combinations in which the number 54 comes at the start  is 2! = 6

Hence the numbers greater than 56,000 = 120 - ( 24+ 6) = 120 - 30 = 90 ways.

We have a cube has 6  faces and we have to colour it with 6 different colours .

Then  the no of ways of colouring =6!=720, but in this we will be getting  many overcountings.

 We have there are 24 ways in which we can orient a cube 72024=30

 Hence the number of distinct ways of colouring a cube with 6 different colours is 72024=30
6!24=306!24=306!24=3072024=30 6!246!24=30

We have a regular tetrahedron has 4 faces and we have to colour it with 4 different colours in 4!=24 ways.

But in this we will be getting many overcountings .

We have there are 12 ways in which we can orient a regular tetrahedron  

Hence the number of distinct ways of colouring a regular tetrahedron  with 4 different colours is 2412=2

The number of selections of r objects from the given n objects is denoted by nCr and we have nCr=n!r!(n−r)! 

Now n ties can be selected from a rack displaying 3n different ties in   3nCn=3n!n!(3n−n)!=3n!n!(2n)!   different ways

1.Arrange all the alphabets in alphabetical order ( E, F , I  , W )

2. 4 alphabets can form 4! words = 24 words.

Since we need an even number , ones place can be occupied by only two numbers 2 and 4 in two ways.

Since repetition is not allowed tens place is occupied by remaining 4 numbers in 4 ways and the hundred's place by 3 ways and thousands place in 2 ways.

Hence total number of even numbers can be formed in 1x2x3x4x2 = 48.

Number of ways of forming 5 digit numbers= 5×4×3×2×1=120

 Number of ways of forming 4 digit numbers= 5×4×3×2=120

Number of ways of forming 3 digit numbers= 5×4×3=60

Number of ways of forming 2 digit numbers=5×4=20

Number of ways of forming 1 digit numbers= 5

Hence the total number of ways = 120+ 120 + 60+ 20+ 5 = 325