3 bids

Question

7 companies bid for 3 different contracts. IN how many ways can the contracts be awarded among these companies, if no firm is to receive more than 2 contracts?

Please show your work

sparky · Answer

looks to me like 336 assuming all te contracts are assigned

(7,3) + (3,2)*(7,2)

Shahzad · Answer

504
if all the firms get one contract 7*6*5=210
if one firm gets 2 contracts then 7*7*6 = 294

210+294=504

Antmavel · Answer

Maybe I am wrong
7^3-3 = 340

For each contract 7 possibilities, then you substract 3 because there are 3 ways to get all the 3 contracts (either company 1, 2 or 3 whereas other companies will get no contracts)

july05 · Answer

the answer is P(7,3)+C(7,2)xP(3,2)=336

could you please explain how you solved it, Sparky?

Shahzad · Answer

i have no idea if i m write or wrong cuz permutation is my weakest area, but first i calculated if companies get one contract atmost then 2 atmost. OA will help infact confirm if i m right or wrong.

july05 · Answer

sorry, the OA is 336. i did not make it clear when i posted the OE

sparky · Answer

Case 1. Firms get 1 contract or none

Two ways to think about it 
1 - number of   combinations   of three firms out of seven - 7!/(3!4!) and then permutate contracts within each pick 3!, => 3! * 7!/(3!4!) = 7!/4!

2 - the above is basically a number of   permutations   of 3 out of 7 (number of arranged picks of 3 out of 7). it is like matching varying arrangemnts of 3 picked firms with a fixed arrangement of 3 contracts 

Case 2. 1 firm gets 2 contracts, another one 1, and the rest none
Same reasoning (think of 2 contracts assigned to a single firm as one contract)
and in addition you multiply what you get by 3!/(2!1!) - number of   combinations  of 2 out of 3 (ordering doesn't matter here since 2 picked contracts are assgined to the same firm)

3!/(2!1!)* 2!*7!/(5!2!)


Total number is Case 1 + Case 2 = 336

christoph · Answer

i correct my answer. it is 336. my solution:

case 1: we have to arrange 7 objects.7!. there are 4 spaces that are equal and 3 different objects. so we have to divide by 4! because the spaces are equal. 7!/4! = 210

case 2: we have to arrange 7 objects. 7!. there are 5 equal spaces and 2 different objects. so we have to divide by 5!. the 2 different objects can be arranged in 3 ways. we multiply by 3. 7!/5! * 3 = 126

210+126=336

ssap · Answer

One more method for Part 2 of the problem (I think we don't hv a problem with Part 1)

No of ways of in which one comapny gets 2 contracts & another company gets 1 contract:


Ways in which one contract(consider this as one unit containing one contract) is awarded to one company = 7
Ways in which the 2 contracts (consider them as one unit containing 2 contracts) can be awarded to one company = 6
..Not stopping here, in how many ways can the 3 contracts be broken into the one contract unit and the 2 contracts unit? =>3C2 (or 3C1) ways = 3

Therefore, no of ways = 7*6*3 = 126

HongHu · Answer

No need to make it this complicated. Antmavel has good grasp on this.

 

Only thing is there are 7 ways to get all 3 contract to one company. Company 1, or 2, or 3, ... or 7 can get all of them. So the correct answer would be 

7^3-7=336

sparky · Answer

nice, no surprise you nailed GMAT with 780

sometimes it really pays to approach a problem from the opposite side!

july05 · Answer

that is definitely ingenious.

Antmavel · Answer

My bad...It is totally logical, I still don't know why I chose 3 instead of 7, nonsense. Thanks HongHu

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