Permutations and Combinations Test - 3

Let the friends be A,B,C,D,E,F,G,H,I , J and assume A and B will not battend together

Case 1 : Both of them will not attend the party : Now we have to select the 6 guests from the remaing 8 members 

 Then no of ways is ⁸C₆ = 28 ways.

Case 2 : Either of them are selected for the party :Now we have to select the 5 guests from the remaing 8 members and one from A and B

Then the no of ways = ²C₁ x ⁸C₅ =112 ways.

Therefore total number of ways is 28+ 112 = 140 ways,

There can either be a heads or tails, therefore for every toss, the possible outcomes are 2. hence for n number of toss the possibilities are 2ⁿ.

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

A team of 4 players are to be selected.

2 out of 6 men can be done in ⁶C₂ ways. 2 out of 4 women can be done in ⁴C₂ways. 

So the number of ways to select 4 players is ⁶C₂x ⁴C₂= 90.

Now we can arrange these people to form mixed doubles.

 If M₁,M₂,W₁,W₂, are the 4 members selected then one team can be chosen as (M₁,W₁)or(M₁,W₂) in 2 different ways 

Therefore the required number of arrangements= 90x2 = 180.