The ABC Society has 4 vice presidents (VPS) and 8 assistant vice-presidents (AVP

Question

The ABC Society has 4 vice presidents (VPS) and 8 assistant vice-presidents (AVPs). Every Wednesdays, the ABC President calls and meets one VP and 4 AVPS for a weekly meeting to discuss improvements that can be done in the organization.

a) One Wednesday, the officers held their meeting at a lunch counter consisting of 10 seats how many seating arrangements are possible if the seats are arranged in a straight line?

b) two VPs are sworn enemies ^ can never sit together while two of the AVPS are sweethearts and would always want to sit together. If all 12 officers excluding the president are to sit in a circular counting of 12 seats, how many seating arrangements are possible? (PLS EXPLAIN THIS ONE I REALLY DONT KNOW :( )

c) In one Saturday, the 12 officers excluding the president decided to have a game to improve the camaraderie a team will consist of 4 members: One VP and 3 AVPs. Thus, 8 are to be selected. From the 8 selected, two competing teams will be formed. In how many ways can the teams be filled.

d) The group of VPS decided to play against the eight AVP. A coin tossed 6 times. If the total number of heads exceeds the total number of tails the VPs will win.

Explanation / Answer

Assuming all VPs and AVPs are treated distinctly, unless specified otherwise,

——

a)

1 President, 1 VP and 4 AVPs

10 seats in straight line => (10C6)*(6!) = 151200 ways [Answer (a)]

—

b)

Take the 2 sweetheart AVPs => Place them together in the circular table arrangement, 2 ways because they can sit on right or left of each other => (12 ways)*(2 ways)

Take one of the two sworn VP enemies - there are 2 cases:

   Case 1: Place adjacent to one of the AVP sweethearts. Then, we can place the other enemy in any other seat NOT adjacent to 1st enemy => (2 ways)*(8 ways)

   OR

   Case 2: Place NOT adjacent to any of the AVP sweethearts Then, we can place the other enemy in any other seat NOT adjacent to 1st enemy => (8 ways)*(7 ways)

This whole exercise of placing the sworn enemies will count cases twice (example: if sworn-enemy #1 is adjacent to sweetheart #1 and sworn-enemy #2 is at position 5; another case could be when sworn-enemy #1 is at position 5 and sworn-enemy #2 is adjacent to sweetheart #1) —> This is fine because we had to consider the position of the sworn-enemies WITHIN the two chosen seats as well => Will NOT multiply by 2 again

Now, place all other 8 people randomly (no constraints) => (8! ways)

=> ANSWER: 12*2*(2*8 + 8*7)*(8!) = 24*72*8! = 69,672,960 seating arrangements are possible [Answer (b)]

—

c)

2 teams: 1 VP and 3 AVPs each

Choose 2 VPs <— (4C2 ways)

Choose 6 AVPs <— (8C6 ways)

Choose 1 VP for team #1 from the 2 chosen VPs <— (2C1 ways)

Choose 3 AVPs for team #1 from the 6 chosen AVPs <— (6C3 ways)

=> Total ways to create teams = (4C2*8C6*2C1*6C3) = 6*28*2*20 = 6,720 ways [Answer (c)]

—

d)

P(number of heads exceeds number of tails in 6 tosses of a fair coin) = P(no. of H > 3) = P(no. of H = 4) + P(no. of H = 5) + P(no. of H = 6)

= (6C4)*(1/2)^6 + (6C1)*(1/2)^6 + (6C0)*(1/2)^6 = (15+6+1)/(2^6) = 22/64 = 11/32 [Answer (d)]

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