A woman has 11 close friends. What is the number of ways she can invit

Question

A woman has 11 close friends. What is the number of ways she can invite 5 of them to dinner if two of the friends are not speaking with each other and will not attend together?

A. 84
B. 126
C. 210
D. 252
E. 378

A woman has 11 close friends. Find the number of ways she can invite 5 of them to dinner where: Two of the friends are not speaking with each other and will not attend together

jimar · Answer

Let her friends be:   and J and K are not on speaking terms, and will not attend together. 
So total possible ways she can invite: 
a) Either invite J or invite K (i.e. one of J and K) + 4 friends except J and K = 2C1 (choose 1 from J and K) * 9C4 (4 of other 9 friends)
b) Don't invite both J and K = 9C5 (choosing 5 friends of rest 9 people)

So total possible invites = 2C1*9C4 + 9C5 
 = 2C1*9C5 + 9C5 (as nCk = nCn-k)
 = 9C5 * (2C1+1) = 378 (E)

Archit3110 · Answer

total possiblities of all guest - including two guests who are not speaking with each other = ways that 2 guests will not attend
11c5-(9c3*2c2) ; 378
IMO E

EgmatQuantExpert · Answer

Solution  
   Given:     
 • A woman has 11 close friends

To find:     
 • What is the number of ways she can invite 5 of them to dinner if two of the friends are not speaking with each other and will not attend together?

Approach and Working Out:     
 • Total ways of selecting 5 out of 11 =  ^{11}C_5 
• Total ways in which 5 are selected, which includes both the friends who do not attend together =  ^2C_2 * ^9C_3

Therefore, the required answer =  ^{11}C_5 – ^2C_2 * ^9C_3 = 378

Hence, the correct answer is Option E.

Answer: E

gvij2017 · Answer

Bunnel!

Is there something wrong with answer?
Even I am getting 378.

Probably, this problem needs a small correction.

Bunuel · Answer

The OA is E. Edited. C is the OA of the following question: https://gmatclub.com/forum/a-woman-has- ... 05760.html

ScottTargetTestPrep · Answer

The number of ways to invite 5 of 11 friends, with no restrictions, is:

11C5 = (11 x 10 x 9 x 8 x 7)/5! = 11 x 3 x 2 x 7 = 462 ways

The number of ways to select the group when the two friends are together is:

2C2 x 9C3 = 1 x (9 x 8 x 7)/3! = 3 x 4 x 7 = 84 ways.

Thus, the number of ways to select the group when the two friends are not together is:

462 - 84 = 378

Answer: E

ThatDudeKnows · Answer

There are some great answers posted above! I just wanted to point out that we can likely save some time if we are comfortable with ballparking and deploying a little logic. There are 11C5 ways to choose the invitees with no restrictions. That's 11*10*9*8*7/5*4*3*2 = 77*2*3 = a little less than 6*80, or 480.

There are four possible scenarios: 
both enemies are invited, 
enemy A is invited but B is not, 
enemy B is invited but A is not, or 
neither enemy is invited.

We need to exclude "both enemies are invited." Is that going to be more than half or less than half of the total possibilities? Less than half. A, B, and C are wrong. D and E are pretty far apart. D would require us to exclude almost half. Is "both enemies are invited" in almost half of the arrangements or far fewer than that? Well, there are 6 people not invited to dinner, so it's more likely to have "neither enemy is invited" than it is to have "both enemies are invited." D is wrong.

Answer choice E.

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A woman has 11 close friends. What is the number of ways she can invit

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A woman has 11 close friends. What is the number of ways she can invit

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