Simple statistics combinations/permutation problem

[Gameowner]

Homework Statement

A touring ‘Blue Caps’ party of 20 cricketers consists of 9 batsmen, 8 bowlers and 3 wicket keepers. A
team of 11 players must be chosen from the players such that there are at least 5 batsmen, 4
bowlers, and 1 wicket keeper. How many different teams can be selected?
(a) if all the players are fit for selection,
(b) if 2 batsmen and 1 bowler are injured and cannot play?

Homework Equations

The Attempt at a Solution

This problem is simple but for some reason I'm stuck...so I guess not easy after all (for me)

We've only practiced with questions that ask a certain a certain combination from a set amount n. The contraints on the last part of that question confuse me, how do we ensure that each amount of those are selected?

Am I just over-reading the question? Is it just as simple as 20C11 on the calculator?

 

[Ray Vickson]

Homework Statement

A touring ‘Blue Caps’ party of 20 cricketers consists of 9 batsmen, 8 bowlers and 3 wicket keepers. A
team of 11 players must be chosen from the players such that there are at least 5 batsmen, 4
bowlers, and 1 wicket keeper. How many different teams can be selected?
(a) if all the players are fit for selection,

(b) if 2 batsmen and 1 bowler are injured and cannot play?

Homework Equations

The Attempt at a Solution

This problem is simple but for some reason I'm stuck...so I guess not easy after all (for me)

We've only practiced with questions that ask a certain a certain combination from a set amount n. The contraints on the last part of that question confuse me, how do we ensure that each amount of those are selected?

Am I just over-reading the question? Is it just as simple as 20C11 on the calculator?

Reply for part (a): If (i,j,k) = numbers of (batsmen, bowlers, keepers) we must have either (i) (6,4,1), (ii) (5,5,1), or(iii) (5,4,2). How many choices do you have in case(i)? In case (ii)? In case (iii)?

RGV

 

[Gameowner]

Reply for part (a): If (i,j,k) = numbers of (batsmen, bowlers, keepers) we must have either (i) (6,4,1), (ii) (5,5,1), or(iii) (5,4,2). How many choices do you have in case(i)? In case (ii)? In case (iii)?

RGV

Since they're independent.

i)24
ii)25
iii)40

so 89?

 

[Ray Vickson]

Since they're independent.

i)24
ii)25
iii)40

so 89?

Show your work.

RGV

 

[Gameowner]

Opps done it slightly wrong, I'll just do i).

(6,4,1) using combinations and multiplicative rule.
= C(9,6)*C(8,4)*C(3,1)
=17,640

Seems a bit wrong...

 

[Ray Vickson]

Opps done it slightly wrong, I'll just do i).

(6,4,1) using combinations and multiplicative rule.
= C(9,6)*C(8,4)*C(3,1)
=17,640

Seems a bit wrong...

It's exactly right.

RGV

 

[Gameowner]

Seemed a bit too large...

So for the total, you would add up all 3 combinations that can generate correct?

and for b), it would be the exact same approach but the totals would decrease by the amounts specified?

Thanks for the help.