Combinations part 2 (1 Viewer)

[Smilebuffalo]

11. An eleven is to be chosen from 15 cricketers of whom 5 are bowlers only, 2 others are wicketkeepers only and the rest are batsman only. How many possible elevens can be chosen which contain at least 4 bowlers and at least 1 wicketkeeper? (answer = 742

18. In how many ways can n things be shared between 2 people? (answer = 2^n -2)

 

[Timothy.Siu]

11. An eleven is to be chosen from 15 cricketers of whom 5 are bowlers only, 2 others are wicketkeepers only and the rest are batsman only. How many possible elevens can be chosen which contain at least 4 bowlers and at least 1 wicketkeeper? (answer = 742

18. In how many ways can n things be shared between 2 people? (answer = 2^n -2)

for 11. u just have to take a few cases, which i'm too lazy to do at this hour.
so, basically write down all the cases.
its the only way i think

18. each "thing" has 2 options. so its like n 2's 2x2x....x2=2^n but this includes the cases where one of them has nothing. so we minus 2 of these cases. 2^n-2

 

[Smilebuffalo]

alright. Thank you very much.

 

[Smilebuffalo]

bump. Q11 anyone?

 

[gurmies]

Case 1 (4 bowl, 1 wick, 6 bat) - 5C4.2C1.8C6 = 280

Case 2 (4 bowl, 2 wick, 5 bat) - 5C4.2C2.8C5 = 280

Case 3 (5 bowl, 1 wick, 5 bat) - 5C5.2C1.8C5 = 112

Case 4 (5 bowl, 2 wick, 4 bat) - 5C5.2C2.8C4 = 70

Sum them up, giving 742.

 

[Smilebuffalo]

Case 1 (4 bowl, 1 wick, 6 bat) - 5C4.2C1.8C6 = 280

Case 2 (4 bowl, 2 wick, 5 bat) - 5C4.2C2.8C5 = 280

Case 3 (5 bowl, 1 wick, 5 bat) - 5C5.2C1.8C5 = 112

Case 4 (5 bowl, 2 wick, 4 bat) - 5C5.2C2.8C4 = 70

Sum them up, giving 742.

Ohhhhhh....


But in the question there's no restriction for the number of batsman that the team can have.

So couldn't we continue and do combinations like:

6 bowl, 2 wick, 3 bat

7 bowl, 2 wick, 2 bat

or....

5 bowl, 3 wick, 3 bat.. etc

 

[lychnobity]

So couldn't we continue and do combinations like:

6 bowl, 2 wick, 3 bat

7 bowl, 2 wick, 2 bat

or....

5 bowl, 3 wick, 3 bat.. etc

But the question says you only have 5 bowlers, and 2 wicketkeepers. That's why you can't have those combinations

 

[Smilebuffalo]

oooh. so out of the 11 chosen, only a maximum 5 can be bowlers and a maximum 2 be wicketkeepers?

 

[Aquawhite]

Ah, good old Combinations again

Perms are just as nice lol... -_- but the ones in the actual HSC MX1 papers are always quite simple.

 

[Smilebuffalo]

haha thats reassuring i guess. I hate perms and combs. It gives me a headache

 

[Thecorey0]

Ah, good old Combinations again

Perms are just as nice lol... -_- but the ones in the actual HSC MX1 papers are always quite simple.

Hopefully that trend continues.

 

[lychnobity]

oooh. so out of the 11 chosen, only a maximum 5 can be bowlers and a maximum 2 be wicketkeepers?

yeah

 

[Aquawhite]

I'm quite confident it will