hi all...im new here...
i wanted to ask is dat in certain questions on combination i see + sign used n in others the * sign used. i.e. 5C4*3C2, or 5C4 + 3C2. is there any particular way to know when to use the + sign or the * sign???? pls help
+ sign mean addition exclusive events
* mean multiply basically dependent events
+ sign mean addition exclusive events
* mean multiply basically dependent events
Thank you 😃
Question :
In how many ways 10 identical presents be distributed among 6 children such that none gets zero.
Ans : I am using the logic
0000000000 Let these be the 10 objects ,Now to divide them i need to put 5 sticks b/w them.
000\000\000\000
this can be done in 9c5 ways . but the answer is 15 c5 . kindly help , where am i going wrong
a+b+c+d+e+f=10
15c5
rajr Sayshow many 4 digit numbers that are divisible by 3,can be formed using the digits 0,1,2,3 and 8 if no digit is to occur more than once in each number?
0+1+2+3.......... 3*3*2*1
0+1+3+8
so total 2*3*3*2 = 36 no.
Q:)In how many ways can we put 7 different balls into 5 different boxes, such that no box is empty?
Why won't the following approach work?
First, we select 5 balls out of 7 in 7C5 = 21 ways.
These 5 balls are distributed in 5 boxes in 5! = 120 ways, now that 5 balls have been put in 5 boxes, one in each, so that no box remains empty.
The remaining 2 balls are put in the 5 boxes in 5X5 = 25 ways.
=>Ans.= 21X120X25 = 63000.
OA is given as 16800.
Please tell me where my approach is wrong.
each 7 ball has 5 choice
5^7 -1 (no box is empty ) = 78124 my take
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?
total arrangement 7!
ABC can be arranged in 6 ways only 1 is desired so 7! /6=840
i am revising the previous mocks and came across a qn( AIMCAT 1206)
i dont remember the exact wording now but it was like this.
6 people have to be distributed in 3 diff cars with atleast 1 in each.
i could solve it till this point.
the no of diff ways are
4,1,1
2,2,2
3,2,1
after this i am not getting the arrangement part. somebody plz
explain .
thanks in advance...
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.
Please help. Thank you.
(4 bowlers, 1 wicketkeeper, 6 batsman)
6C4*2c1*10c6 = 6300
(4 bowlers, 2 wicketkeeper, 5 batsman)
6C4*1*10c5= 3780
(5 bowlers, 2 wicketkeeper, 4 batsman)
(5 bowlers, 1 wicketkeeper, 5 batsman)
(6 bowlers, 2 wicketkeeper, 3 batsman)
(6 bowlers, 1 wicketkeeper, 4 batsman)
ankit13 Saysthanks...can u plz explain the concept of distributing different to similar and different to different with some conditions
(a)DIFFERENT TO DIFFERENT:
You have to select the things as well as groups, as each selection will be different since the groups are not identical.
Eg:Distribute 1,2,3 amongst A,B.
Possibilities:
A_____________B
1_____________2,3
2_____________1,3
3_____________1,2
1,2____________3
1,3____________2
2,3____________1
X_____________1,2,3
1,2,3__________X
No. of ways can be calcuted as this:
Let A get 2 things and B will automatically get 1 thing. Since things are different, select 2 things for A in 3C2 = 3 ways, and 1 thing out of 1 in 1C1 = 1 way, so total 3X1 = 3 ways. (cases 4,5,6 in my example). Similarly, if A gets 1 thing, you just repeat the process and add the total no. of cases. Also, there will be 2 cases when all things go in 1 group only.
Had there been 3 groups, you would have had to consider all the possible cases.(1,1,1;1,2,0..etc.).
(B)DIFFERENT TO SIMILAR:
Again, you have to select the things, since they are all distinct and will make different selections; but you don't have to select the groups.
So, consider now the following cases from my example:
1_________2,3
2_________1,3
3_________1,2
X_________1,2,3
There will be NO more cases, because, since the groups are similar, it will just be a repetition. It's like putting different coloured caps in identical saucers(imagine it, it will make it easier to understand).
The main difference in thinking will be this:
Let A get 2 things so B will get 1. Select 2 things in 3C2 = 3 ways.
You don't have to count these 3 cases again (as we did earlier).
Hope it is clear now.
i have a doubt;
there was a question somewhere where there are mXn cells, and u hav to find the total number of ways to go from one point say a (i1, j1) to b(i2, j2). ....
please help
i have a doubt;
there was a question somewhere where there are mXn cells, and u hav to find the total number of ways to go from one point say a (i1, j1) to b(i2, j2). in the shortest possible way ....
please help
Q39.How many different nine digit numbers can be formed from the number
223355888 by rearranging its digits so that the odd digits occupy even positions?
(a) 16(b) 36 (c) 60
q39.how many different nine digit numbers can be formed from the number
223355888 by rearranging its digits so that the odd digits occupy even positions?
(a) 16(b) 36 (c) 60
4!/(2!2!)* 5!/(2!3!)=60
Q39.How many different nine digit numbers can be formed from the number
223355888 by rearranging its digits so that the odd digits occupy even positions?
(a) 16(b) 36 (c) 60
My take is 60.
_ _ _ _ _ _ _ _ _
=> 4 even places to be filled with (3, 3, 5, 5) => 4!/(2!*2!) ways = 6 ways
=> 5 odd places to be filled with (2, 2, 8, 8, 8 ) => 5!/(3!*2!) ways = 10 ways
Total 6*10 = 60 ways
Q38.A five-digit numbers divisible by 3 is to be formed using the numerals 0, 1, 2,
3, 4 and 5, without repetition. The total number of ways this can be done is
(a) 216 (b) 240
(c) 600 (d) 3125
Q38.A five-digit numbers divisible by 3 is to be formed using the numerals 0, 1, 2,
3, 4 and 5, without repetition. The total number of ways this can be done is
(a) 216 (b) 240
(c) 600 (d) 3125
is it (a) 216
please help....
In how many ways 6 men and 6 women can be seated in a row such that 2 particular men are always together and 2 particular women are never together..
please help....
In how many ways 6 men and 6 women can be seated in a row such that 2 particular men are always together and 2 particular women are never together..
let me know if I am wrong
,2*11! - 4*10!
which would be,
(2*10!)(11-2) or 18*10!
please help....
In how many ways 6 men and 6 women can be seated in a row such that 2 particular men are always together and 2 particular women are never together..
As my understanding goes if 6 men are named from M1, M2.... M6, then how many seating arrangement is possible if M1, M2 are together similarly for other combination!!!
If M1, M2 are together you can arrange them in
5! ways x 2
and girls in between chairs of men in 6! ways !!
so in total 5!x2x6!
combination of men possible is 6c2 ways =>15
=>5!x2x6!x15