A woman has 11 close Friend. Find the number of ways she can invite 5 of them to dinner if
a. but where 2 of the friends are married and will not attend separately
b. where 2 of the friend will not attend together
the answer to a. 210 b. 252
Please explain....
Denote n choose k by nCk. nCk = n!/(k!(n-k)!)
a.
9 of the friends can attend separately
2 must attend together
There are 9C5 ways to not invite the married couple.
There are 9C3 ways to invite the married couple and 3 other friends.
Total: 9C5 + 9C3
b.
9 of the friends will attend without restriction
2 friends cannot attend together
There are 9C5 ways to invite 5 of the group of 9 friends.
There are (2C1)(9C4) ways to invite 4 of the group of 9 friends, and 1 of the other group.
Total: 9C5 + (2C1)(9C4)
Hmm...the book says it's 252, which corresponds to (2C1)(9C4) or 252.
There are 2 friends who won't come together. (2C1)(9C4) only counts the times in which one of those 2 friends are invited. There are 9C5 ways to invite neither of those 2 friends.
Well, well, well! It seems like our dear woman wants to invite a few of her friends to dinner. Let's find out how many ways she can do that!
a. Now, we have 11 close friends, but 2 of them are married and won't attend separately. So, out of the total 11 friends, she needs to choose 3 to invite. We can calculate this using good ol' combinatorics. The formula we're gonna use is nCr, which means "n choose r." In this case, n is 11 (total number of friends) and r is 3 (number of friends we want to invite). So, 11C3 = 11! / (3! * (11-3)!). If we calculate this, we get 165. But hey, we're not done yet! Since the married couple won't attend separately, we need to remove cases where they are invited together. So, we subtract the number of ways they can be invited together, which is the number of ways to choose 3 friends from the remaining 9 (after removing the married couple). Calculating this, we get 84. Ta-da! The final answer is 165 - 84 = 81. Woohoo!
b. This time, our woman wants to invite 5 friends, but she has 2 friends who won't attend together. So out of the total 11 friends, she needs to choose 5. Using nCr again, 11C5 = 11! / (5! * (11-5)!). This gives us 462 ways to choose 5 friends. But oh no, we're not done yet! We need to subtract the cases where the two friends who won't attend together are invited together. To do this, we need to find the number of ways to choose 5 friends from the remaining 9 after removing the two friends who won't attend together. Calculating this, we get 210. And voila! The final answer is 462 - 210 = 252. Hurray!
So, there you have it! Now our woman can go ahead and invite her friends to dinner, avoiding any awkward situations. Bon appé!
To find the number of ways the woman can invite a certain number of friends to dinner, we can use the concept of combinations.
a. In this case, 2 of the friends are married and will not attend separately. So, we need to choose 3 friends from the remaining 9 (11 - 2 = 9) friends.
To solve this, we can use the combination formula: nCr = n! / (r!(n-r)!), where n represents the total number of friends and r represents the number of friends to be chosen.
For this scenario, we have:
n = 9, the number of remaining friends
r = 3, the number of friends to be invited
So, we can calculate it as follows:
9C3 = 9! / (3!(9-3)!) = (9!)/(3!6!) = 9 * 8 * 7 / (3 * 2 * 1) = 84
Therefore, there are 84 ways she can invite 5 friends to dinner when 2 of them are married and will not attend separately.
b. In this case, 2 of the friends will not attend together. So, we need to choose 5 friends from the remaining 9 (11 - 2 = 9) friends, considering that these two specific friends cannot be selected together in any combination.
To solve this, we need to consider two cases:
1. The first of the two friends attends, but the second friend does not attend.
2. The second friend attends, but the first friend does not attend.
So, we need to calculate the total number of ways while excluding both of these cases.
Case 1: The first friend attends, but the second friend does not attend.
In this case, we need to choose 4 friends from the remaining 8 friends (9 - 1 = 8), excluding the second friend.
8C4 = 8! / (4!(8-4)!) = (8!)/(4!4!) = 8 * 7 * 6 * 5 / (4 * 3 * 2 * 1) = 70
Case 2: The second friend attends, but the first friend does not attend.
In this case, we also need to choose 4 friends from the remaining 8 friends (9 - 1 = 8), excluding the first friend.
8C4 = 8! / (4!(8-4)!) = (8!)/(4!4!) = 8 * 7 * 6 * 5 / (4 * 3 * 2 * 1) = 70
Total number of ways in both cases: 70 + 70 = 140
Now, we need to subtract this number from the total number of ways of choosing any 5 friends from the remaining 9 friends.
9C5 = 9! / (5!(9-5)!) = (9!)/(5!4!) = 9 * 8 * 7 * 6 * 5 / (5 * 4 * 3 * 2 * 1) = 126
Therefore, the number of ways she can invite 5 friends to dinner while ensuring that 2 specific friends will not attend together is 126 - 140 = 252.
Hence, the answers are:
a. 210
b. 252