I solved it this way but am mot 100% sure.. MATHEMATICS has 2M , 2 A , 2T and 5 other letters.Three cases arise: case1- when all letters are distinct. E has to be there and remaining four letters can be selected in 7C4 ways. So, 7C4*5! =1050 Case2- when 2 pair of letters are alike (ie AAETT) (3C2*5!)/2!*2! =90 (3C2 for selecting 2 letters from M,A,T) case3-when 1 pair of letter is alike (ie AAEIS or ETTIC) (3C1*6C2*5!)/2!=2700 But I am getting total 3840 ways only
Hello Puys, Kindly have a look at the problem below:
Question A cricket team of 11 players is to be selected from 18 players, of whom 10 are batsmen, 6 are bowlers and 2 are wicketkeepers. If the team should have at least 4 boowlers and atleast 1 wicketkeeper, the number of ways in which the team can be formed is?
A) 15000 B) 14550 C) 17850 D) 14904 E) 15800
Now, the thing is that I know the answer to this question which is determind by making different situations on the number of bowlers, batsmen & WC as follows and then making combinations: BAT BOWL WC 6 ..... 4 .... 1 5 ..... 5 .... 1 4 ..... 6 .... 1 5 ..... 4 .... 2 4 ..... 5 .... 2 3 ..... 6 .... 2
Now please look here, why cant we solve this problem as follows: (Selecting 4 bowlers out of 6) x (Selecting 1 WC out of 2) X (Selecting 6 players out of remaining 13 players) Please tell me the problem with the above logic, I know it's wrong but can not figure out as to why it is wrong.
Hello Puys, Kindly have a look at the problem below:
Question A cricket team of 11 players is to be selected from 18 players, of whom 10 are batsmen, 6 are bowlers and 2 are wicketkeepers. If the team should have at least 4 boowlers and atleast 1 wicketkeeper, the number of ways in which the team can be formed is?
A) 15000 B) 14550 C) 17850 D) 14904 E) 15800
Now, the thing is that I know the answer to this question which is determind by making different situations on the number of bowlers, batsmen & WC as follows and then making combinations: BAT BOWL WC 6 ..... 4 .... 1 5 ..... 5 .... 1 4 ..... 6 .... 1 5 ..... 4 .... 2 4 ..... 5 .... 2 3 ..... 6 .... 2
Now please look here, why cant we solve this problem as follows: (Selecting 4 bowlers out of 6) x (Selecting 1 WC out of 2) X (Selecting 6 players out of remaining 13 players) Please tell me the problem with the above logic, I know it's wrong but can not figure out as to why it is wrong.
Question A cricket team of 11 players is to be selected from 18 players, of whom 10 are batsmen, 6 are bowlers and 2 are wicketkeepers. If the team should have at least 4 boowlers and atleast 1 wicketkeeper, the number of ways in which the team can be formed is?
Now please look here, why cant we solve this problem as follows: (Selecting 4 bowlers out of 6) x (Selecting 1 WC out of 2) X (Selecting 6 players out of remaining 13 players)
We can not do it like this, as there is lot of overcounting. Consider following ywo cases:-
i) You choose b1, b2, b3, b4, then w1 and then choose b5, B1, B2, B3, B4, B5 ii) You choose b2, b3, b4, b5, then w1 and then choose b1, B1, B2, B3, B4, B5
Both the cases are different but you counting them same. Similarly there are many repetitions
We can not do it like this, as there is lot of overcounting. Consider following ywo cases:-
i) You choose b1, b2, b3, b4, then w1 and then choose b5, B1, B2, B3, B4, B5 ii) You choose b2, b3, b4, b5, then w1 and then choose b1, B1, B2, B3, B4, B5
Both the cases are different but you counting them same. Similarly there are many repetitions
Oh Yes, You're right.. Hmm.. Silly me.. :shocked: Thank You
Sorry for this silly que bt m nt getting the answer given in the book ... The no. of ways that 7 person can address a meeting so that three persons A,B and C from them A will speak before B and B before C is? 840 or 720?
Sorry for this silly que bt m nt getting the answer given in the book ... The no. of ways that 7 person can address a meeting so that three persons A,B and C from them A will speak before B and B before C is? 840 or 720?
My take is 840.
Maximum arrangements = 7!. But from A, B and C; out of 3! arranagements only 1 is what we need. So, out of 7!; we will need 7!/3! = 840.
Sorry for this silly que bt m nt getting the answer given in the book ... The no. of ways that 7 person can address a meeting so that three persons A,B and C from them A will speak before B and B before C is? 840 or 720?
x.....A.....y.....B.....z.....C.....s
x+y+z+s = 7-3 = 4
Solutions : C(7,3) : 35.
But the remaining can also be permuted in 4! = 24 ways
The probability of a missile hitting a target is 1/2 and it takes 3 missiles to destroy a bridge, atleast how many missiles should be fired such that the probability of destroying the bridge is greater than 0.999
6 people have to go to attend a meeting in 3 different vehicles, each of which can accommodate 6 persons. In how many ways they can go so that they use all the three vehicles?
