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# A man has three friends. The number of ways he can invite one friend everyday for dinner on six successive nights so that no friend is invited more than three times is-${\text{A}}{\text{. 640}} \\ {\text{B}}{\text{. 320}} \\ {\text{C}}{\text{. 420}} \\ {\text{D}}{\text{. 510}} \\$

Last updated date: 15th Sep 2024
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Hint: Guess the patterns possible for inviting the three friends. After this calculate the respective combinations. The sum of all the cases will be the answer.

The possible patterns of inviting 3 friends such that no one is invited more than 3 times, s per the number of invitations, will be-
(3,3,0), (2,2,2), (1,2,3)
Where each friend has a maximum occurrence of 3.
Solving for the first case (3,3,0),
Here any 2 friends get 3 turns to visit.
Selecting 2 friends from 3 - ${}^3{C_2}$
Selecting days - ${}^6{C_3} \times {}^3{C_3}$
Becomes- ${}^6{C_3} \times {}^3{C_3} \times {}^3{C_2} = 60$
Solving for the second case - (2,2,2)
Each friend is invited equally
Selecting friend - ${}^3{C_3}$
Selecting days - ${}^6{C_2} \times {}^4{C_2} \times {}^2{C_2}$
Becomes - ${}^6{C_2} \times {}^4{C_2} \times {}^2{C_2} \times {}^3{C_3} = 90$
Solving for the third case - (1,2,3)
Here, we have all three friends. Now, any one of them can get 1 turn or 2 turns or 3 turns. So here, there is a need for an arrangement.
Selecting and arranging friends - ${}^3{C_3} \times 3!$
Selecting days - ${}^6{C_1} \times {}^5{C_2} \times {}^3{C_3}$
Becomes - ${}^6{C_1} \times {}^5{C_2} \times {}^3{C_3} \times 3! = 360$
The final answer becomes the sum of all the three cases-
360+90+60 = 510
The correct option is D.

Note: In this question, the determination of the different patterns possible is the important step. And the next notice goes to the third case (1,2,3) where we rearranged, as this was not unique to any one friend.