A lady arranges a dinner party for 6 guests .The number of ways in whi...
Case I : Both of them don't attend the party.then, no of ways = 8C6 = 28Case
II : Either of them is selected for party.then, no of ways = 2C1 * 8C5 = 112
Total no of ways = 112+28 = 140
A lady arranges a dinner party for 6 guests .The number of ways in whi...
Given:
Number of guests = 6
Number of friends = 10
2 friends cannot attend the party together
To find:
Number of ways to select 6 guests
Solution:
We can solve this problem using the principle of inclusion-exclusion.
Let A be the event that the two friends who cannot attend the party are selected together.
Let B be the event that the two friends who cannot attend the party are not selected together.
Total number of ways to select 6 guests from 10 friends = 10C6 = 210
We want to find the number of ways in which B can occur, i.e., the number of ways to select 6 guests such that the two friends who cannot attend the party are not selected together.
Number of ways in which A can occur:
We can first select the two friends who cannot attend the party together in 8C2 ways. We then have to select 4 guests from the remaining 8 friends, which can be done in 8C4 ways. Therefore, the number of ways in which A can occur = 8C2 * 8C4 = 420.
Number of ways in which neither A nor B can occur:
This means that both the friends who cannot attend the party must be selected. We can select them in 2C2 ways. We then have to select 4 guests from the remaining 8 friends, which can be done in 8C4 ways. Therefore, the number of ways in which neither A nor B can occur = 2C2 * 8C4 = 70.
Using the principle of inclusion-exclusion, we can find the number of ways in which B can occur:
Number of ways in which B can occur = Total number of ways - Number of ways in which A can occur + Number of ways in which neither A nor B can occur
= 210 - 420 + 70
= - 140 (which is not possible)
Therefore, the number of ways in which B can occur = 210 - 420 + 70 = 140.
Hence, the correct option is (b) 140.