39this is a typical tyoe of question in permutation and combination.remember it and it will be useful in many cases.
A cricket eleven has to be chosen from 15 players of whom 6 can bowl and 3 keep wickets,while none of them can keep wicket and bowl.In how many ways can this be done if it is to contain at least 4 bowlers and 2 wicketkeepers
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18 Answers
62the mistake si that u are recounting...
let us see how.. let the two bowlers be B1 and B2
First take 1 bowler
B1.. then take 1 more player .. it contains the case when B2 is taken..
B2.. then take 1 more player .. it contains the case when B1 is taken..
did u get the repetition? ?
Once u have picked some bowlers.. then u should not pick from bowlers again!
62I wud not use the word "Learn"
but i think u should try to udnerstand.. this is indeed a very typical type..
But these are some of those dirty your hand kind of problems which are more common in CBSE 6 marks than in IIT. But still with the objective papers this is very much a possibility!
3962Dear.. this is 960 i solved it after ur post!!
See abhishek's method.. the 6 cases he used
1. 4+2+5
2 5+2+4
3. 6+2+3
4. 4+3+4
5. 5+3+3
6. 6+3+2
case 1=270 ways
case 2=270 ways
case 3=060 ways
case 4=225 ways
case 5=120 ways
case 6=015 ways
The sum is 960!
62Forget that case.. we already have recounted the way a 2nd bowler can be there.. it is in both the counting..
so having 2 bowlers wala case is being counted 2wice.. hence ur answer to this problem is different by "1"
1aapka q says it shud hv @least 1 bowler so it can hav 2 bowlers na...
33One approach can be..
1. 4+2+5
2 5+2+4
3. 6+2+3
4. 4+3+4
5. 5+3+3
6. 6+3+2
4+2+5 means 4 ballers 2 WK and rest 5
so for 1st its 6C4*3C2*6C5
Similarly for 2nd 6C5*3C2*6C4
........
........
Now add all these...
1ya silly me i by mistakenly deletd ma above post sorry
and ya by mymethod m gettin 20 ...........
kewl ....bt hwz dat possible?
wats d mistake in ma logic?
62post using ur own logic...
That answr(20) should be different.. cos i got ur logic wrong !
I guess u can try by ur own logic.. and check if the answer matches 19!
If it does then it means we have an alternate method right..
otherwise something is wrong!
I know what is.. but i want u to think first
62Actual solution:
case 1: 1 bowler: 2. 9 ways
Case 2: 2 bowlers: 1 way
total 19 ways..
62out of 11 players, 2 can bowl
Find the number of ways to pick 2 players such that atleast 1 bowler
@integration:
ur logic will give
pick 1 bowler and then any 1 player .. this can be done is 2.10 ways = 20
Actual solution:
case 1: 1 bowler: 2. 9 ways
Case 2: 2 bowlers: 1 way
total 19 ways..
do u realise where we are recounting?
62even i had first though of this...
oops edited*
6C4 × 3C2 × 15-6C5
This is wrong
For the same reason why ur soln is wrong!
1can we do it in a manner like
we need atleast 4 bowler's and 2 wicket keeper's so we take them and keep them aside
nw we gottu choose frm 9 othr players of wich 1 is a wicketkeepr and 2 r bowlers and we need 2 choose 5 p'pl more
so we can do dat in 3 ways
basically
1) choose both d bowlers and othr 3 playes can b choosen in 7C3 ways
2) choose d wicket keepr and othr 4 can b choosen in 8C4 ways
3)choosen one bowler and one wicketkeepr and otr i similer amnnr as bove
correct me if m wrng :'(