A meeting has to be conducted with 5 managers. Find the

Question

A meeting has to be conducted with 5 managers. Find the number of ways in which the managers be selected from among 9 managers, if 2 managers will not attend the meeting together?

A. 35
B. 91
C. 120
D. 126
E. 150

Bunuel · Answer

C^5_9-C^2_2*C^3_7=91 , where:
 C^5_9  is the total # of ways to choose 5 people out of 9 without any restriction;
 C^2_2*C^3_7  is the # of ways to choose 2 managers who will not attend the meeting together and other 3 managers out 7 left (restriction).

Answer: B.

DriftingAway · Answer

Could someone explain to me what  C^5_9  is? I'm not familiar with anything that takes this form.

I got the right answer by doing 5! (5x4x3x2) and then knowing it had to be less than that because of the restriction but I don't understand what this solution means to do for getting the exact answer.

Bunuel · Answer

C^5_9=\frac{9!}{(9-5)!*5!}=\frac{9!}{4!*5!} is the # of ways to choose 5 items out of 9 distinct items when the order of the selection does not matter. Check this for more: math-combinatorics-87345.html Hope it helps.

akrish1982 · Answer

i did this the other way around

we can either choose all 5 people from 7 manager who have no problems or choose 4 from the 7 and 1 from the 2 managers who have a problem sitting together

so 7C5 + (7C4 * 2C1)

this is 21 + 70

91

Fdambro294 · Answer

(Total No. of Ways to choose ANY 5 with NO Restriction) - (No. of Ways to Choose a Team where 2 MEN are ALWAYS Selected) = No. of Ways to Choose Team in which 2 Men are NEVER on the Same Team

Place the 2 Men on the Team for each Possible Combination (only have to Select 3 now) and Reduce the Overall Pool Available by -2 also.

"9 choose 5" - "7 choose 3" = 9! / (5!4!) - 7! / (4!3!) = 126 - 35 = 91 Ways

-B-

MathRevolution · Answer

Total managers: 9

To be selected: 5

Number of ways:  ^9{C_5}  = 126

If '2' managers always attend meeting together, then we will require only 3 managers to select out of '7':  ^7{C_3}  = 35

Number of ways in which '2' managers will not attend the meeting together: 126 - 35 = 91

Answer B

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A meeting has to be conducted with 5 managers. Find the

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