Robin and Terry want to invite 5 of their friends to their wedding. Ro

Question

Robin and Terry want to invite 5 of their friends to their wedding. Robin has 7 friends, Terry has 6, and Robin and Terry have no friends in common. If at least 1 of Robin’s friends and at least 1 of Terry’s friends must be invited, how many different groups of friends could Robin and Terry invite to their wedding?

A. 462
B. 924
C. 1,260
D. 2,520
E. 151,200

A. 462
B. 924
C. 1,260
D. 2,520
E. 151,200

Archit3110 · Answer

total friends = 13
so 13*12*11*10*9 / 5! ; 1287 groups possible 
now since question has asked at least 1 friend each of tony & robin
Robin 7*6*5*4*3/5! = 21
and Terry ; 6*5*4*3*2/5! = 6
total Robin & Terry = 27
subtract it from 1287 -27 = 1260 groups can be invited

Shobhit7 · Answer

Without restriction: 13C5= 1287
Only Robin's: 7C5= 21
Only Terry's: 6C5= 6

Remaining options: 1287-21-6= 1260

Ans C

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ScottTargetTestPrep · Answer

We see that there are 4 cases possible: Robin invites 1, 2, 3 or 4 friends (notice that Robin can’t invite 0 friend since she has to invite at least 1 friend, and she can’t invite 5 friends since Terry has to invite at least 1 friend).

1) If Robin invites 1 friend, then Terry invites 4, and there are 7C1 x 6C4 = 7 x 15 = 105 ways to invite 5 of their friends.

2) If Robin invites 2 friends, then Terry invites 3, and there are 7C2 x 6C3 = 21 x 20 = 420 ways to invite 5 of their friends.

3) If Robin invites 3 friends, then Terry invites 2, and there are 7C3 x 6C2 = 35 x 15 = 525 ways to invite 5 of their friends.

4) If Robin invites 4 friends, then Terry invites 1, and there are 7C4 x 6C1 = 35 x 6 = 210 ways to invite 5 of their friends.

Therefore, the total number of ways to invite 5 of their friends is 105 + 420 + 525 + 210 = 1,260.
 
Alternate Solution:

If there were no restrictions, 5 people among a total of 6 + 7 = 13 people could have been chosen in 13C5 = 13!/(5!*8!) = (13 x 12 x 11 x 10 x 9)/(5 x 4 x 3 x 2) = 13 x 11 x 9 = 1287 ways.

From these 1287, we must subtract the sum of the number of ways in which Robin invites 5 friends and Terry invites 5 friends.

Since Robin has 7 friends, she can choose 5 friends in 7C5 = 21 ways. Similarly, since Terry has 6 friends, he can choose 5 friends in 6C5 = 6 ways. Therefore, 21 + 6 = 27 of the 1287 choices are those in which either party invites no friends. Thus, 5 friends can be invited in 1287 - 27 = 1260 ways such that either party invites at least one friend.

Answer: C

Novice90 · Answer

What is wrong with this method:

# of ways of selecting one friend of Robin = 7C1 = 7
# of ways of selecting one friend of Terry = 6C1 = 6
# of ways of selecting other 3 friends from the remaining 11 friends= 11C3 = 165

Hence 7x6x165 = 6930

exc4libur · Answer

R: 7 friends, T: 6 friends

: 7C4•6C1 = 210
 : 7C3•6C2 = 525
 : 6C4•7C1 = 105
 : 6C3•7C2 = 420

Total cases (OR=ADD): 210+525+105+420=1260

Ans (C)

mohit2492 · Answer

Consider friends labeled as R1, R2, R3, R4, R5, R6, R7 and T1, T2, T3, T4, T5, T6.

Then first part of your formula "7*6" means that you're choosing one of R friends and one of T friends. Let's say, you've chosen R4 and T2:

{R4, T2, ..., ..., ...}.

Then you're filling out other 3 places by any combination of 3 people among other friends. Let's say, you have chosen R5, T4 and R1:

{R4, T2, R5, T4, R1}.

Now let's start again and say, that in the beginning, we have chosen R5 and T4:

{R5, T4, ..., ..., ...}.

But after we fortuitously can choose R4, T2, and R1. Which gives

{R5, T4, R4, T2, R1}

and we count the same combination for the second time!

ScottTargetTestPrep · Answer

Without any restrictions, 5 people can be chosen from a total of 6 + 7 = 13 people in 13C5 = 13!/(5!*8!) = (13 x 12 x 11 x 10 x 9)/(5 x 4 x 3 x 2) = 13 x 11 x 9 = 1287 ways.

This number includes the cases where all 5 people are Robin’s friends and all 5 people are Terry’s friends; therefore, we must subtract those from the total number, as shown below.

The number of ways 5 people can be chosen from 7 of Robin’s friends is 7C5 = 7!/(5!*2!) = (7 x 6)/2 = 21.

The number of ways 5 people can be chosen from 6 of Terry’s friends is 6C5 = 6C1 = 6 (since it is the same thing as choosing the friend which does not get invited).

Thus, there are 1287 - (21 + 6) = 1260 ways 5 people can be invited to the wedding where at least one friend of each party is invited.

Answer: C

bumpbot · Answer

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Robin and Terry want to invite 5 of their friends to their wedding. Ro

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