Permutations and Combinations Test - 2

He can invite any one  friend in ⁶C₁ ways= 6 ways:

He can invite any two friends in ⁶C₂ ways = 15 ways

He can invite any three friends in ⁶C₃ ways =20 ways

He can invite any 4 friends in ⁶C₄ways = 15 ways

He can invite any 5 friends in ⁶C₅ ways = 6 ways

He can invite  all the 6 friends in ⁶C₆ ways= 1 way.

Since any one of these could happen total possibilities are, 6+15+20+15+6+1 = 63.

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,

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 M_1,M_2,W_1,W_2, 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.

Case 1: for 5 digit numbers Ten Tens thousands place can be occupied in 4 ways since 0 cannot be suitable. For ones place remaining 4 numbers ( since repetition is not allowed ) can be occupied in 4 ways : and the tens place in 3 ways and hundreds place in 2 ways and the thousands place in 1 way. Hence the total number of ways is 4x1x4x2x3= 96.

Case 2: for 4 digit numbers thousands place can be occupied in 4 ways since 0 cannot be suitable. For ones place remaining 4 numbers ( since repetition is not allowed ) can be occupied in 4 ways : and the tens place in 3 ways and hundreds place in 2 ways . Hence the total number of ways is 4x4x2x3= 96.

Case 3: for 3digit numbers hundreds place can be occupied in 4 ways since 0 cannot be suitable. For ones place remaining 4 numbers ( since repetition is not allowed ) can be occupied in 4 ways : and the tens place in 3 ways  Hence the total number of ways is 4x4x3= 48

Case 4: for 2 digit numbers Tens place can be occupied in 4 ways since 0 cannot be suitable. For ones place remaining 4 numbers ( since repetition is not allowed ) can be occupied. Hence the total number of ways is 4x4=16.

Case 5: For single digit in 4 ways .

Hence 96 + 96 + 48+ 16+4 = 260

One's place can be occupied by any of the  5 numbers, tens place by any of the 5 numbers and hundreds place in 5 ways since repetition is allowed. But thousands place can be occupied by 2,3,1, only since the required number should be greater than 1000 and less than 5000.Hence  total number of arrangement=3x5x5x5=375

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.

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.

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.