The remaining 3 digits can be filled in the remaining three places in 3! ways. Hence, the digits at the ten’s place can be 1, 2 or 3. However, if digit 4 is placed at the unit place then 5 cannot occupy the ten’s place. Hence, if 5 is placed at the unit place, then remaining four places can be filled with any of theįour digits 1, 2, 3 and 4, hence total numbers =4! In how many ways can he invite them, if there have to be exactly 3girls in the invitees?Īccording to the question, the digit in the unit’s place should be greater than that in the ten’s place. Question 2: A man has 9 friends, 4 boys and 5 girls. The sum of all the numbers formed by the digits a 1, a 2, a 3,……….a n, without repetition of the digits is given by: Thousand’s place and ten thousand’s place will be 600. Similarly, the sum of the digits at the ten’s place, hundred’s place, Similarly, there will be 24 numbers each with 3, 5, 7 and 9 at the unit’s place.
So, there are 24 numbers which have 1 at the unit’s place and sum of these numbers is 24. If we fix 1 at the unit’s place, the other digits can be arranged in 4! = 24 ways.
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