First we will fix any one digit in a fixed position .Then we have the remaining 4 digits can be arranged in 4! different ways.
Which means each of the five digits can appear in each of the five places in 4! times.
Hence the sum of the digits in each position is
Now to find the sum of these numbers formed we have to consider the place values for these digits
So the sum of all the numbers which can be formed by using the digits 1 , 3 , 5 , 7 ,9=
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 M1,M2,W1,W2, are the 4 members selected then one team can be chosen as 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.
t th | th | h | t | o |
4 | 1 | 2 | 3 | 4 |
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.
th | h | t | o |
4 | 2 | 3 | 4 |
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
th | h | t | o |
3 | 5 | 5 | 5 |
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