@ayushbhalotia said: @rajatt1557 - in the questn 3 should be followed by 6 means 36 or 63...the no of cases change accordingly...
pls give the detailed sol.
In how many ways 3 books can be selected from a set of 10 books which are kept on shelf such that no two books which are selected are adjacent.
Ans: 10C3-56-8
I don't get how are we getting 56? Can any one please help me with this?
@sanjay04ag said: In how many ways 3 books can be selected from a set of 10 books which are kept on shelf such that no two books which are selected are adjacent.Ans: 10C3-56-8I don't get how are we getting 56? Can any one please help me with this?
suppose 10 books are
* * * * * * * * * *
So no of ways of selecting any 3 book from this are 10C3
Now the cases of selecting 3 book adjacent are 8C1=8
Now no of ways of selecting 2 books adjacent are 9*8=72
but in the last case you are considering 3 book adjacent also so we have to subtract those
(* *) * * * * * * * * You have selected 2 books but you may select 3rd book also so u have to subtract 1 for it.
* (* *) * * * * * * * You have selected 2 books but you may select 1st and 3rd book also so u have to subtract 2 for it
* * * * * * * * (* *) You have selected 2 books but you may select 8th book also so u have to subtract 1 for it.
so total no of subtraction is 2*1+7*2=16
so no of ways for selecting 2 adjacent books are are 72-16=56
Total = 10C3-56-8
@rajatt1557 said: How many four digit od numbers can be formed such that every 3 in the no. is followed by 6? a)108 b)2592 c)2696 d)2700
Hi according to me followed means it should be 36 also 63 then ans will be b>2696 otherwise I am getting 2660.....
@[361226:ayushbhalotia] can u explain how r u getting 9 casses.
I am getting only 8.
R=2 : 1 2 4, 2 4 8 ,3 6 12,4 8 16,5 10 20.
R=3 : 1 3 9,2 6 18
R=4 : 1 4 16
total 8 cases.....also if we consider R
@wontmissthstime said: @ayushbhalotia can u explain how r u getting 9 casses.I am getting only 8.R=2 : 1 2 4, 2 4 8 ,3 6 12,4 8 16,5 10 20. R=3 : 1 3 9,2 6 18 R=4 : 1 4 16 total 8 cases.....also if we consider R
Ya i am getting 8 too
4,6,9 ... r = 3/2...
hey puys....
Can we start posting question with a question no. so that it becomes easy to trase questions and their proposed solutions...
@AsihekAdhvaryu
"Now no of ways of selecting 2 books adjacent are 9*8=72"
what i think is no of ways to select 2 adjacent books is only 9 ....how comes 9*8???
plz explain
"Now no of ways of selecting 2 books adjacent are 9*8=72"
what i think is no of ways to select 2 adjacent books is only 9 ....how comes 9*8???
plz explain
how????
@Kapil2012 said: how to select 6 characters from 5 As 4 Bs 3Cs and 2Ds ??
@AsihekAdhvaryu said:Is it 15?
Isn't this question to be done considering the possibility as below:
1st case: Pick 5As n place any of the other three letter as 6th letter=> 1*3= 3 possibilities
2nd case: Pick 4As from 5 and place the remaining two with any of the other letter=> 5c4*4c2(2 Bs from 4) + 5c4*3c2(2 Cs from 3)+ 5c4*2c2( 2 Ds) + 5c4*3(BC, CD, BD)= 65 possibilities..
and so on.. this seems too lengthy.. just tell me what's the right approach
@wontmissthstime said: @AsihekAdhvaryu"Now no of ways of selecting 2 books adjacent are 9*8=72"what i think is no of ways to select 2 adjacent books is only 9 ....how comes 9*8???plz explain
Yesh.. Dr has to be just 9 ways. (1,2)(2,3)(3,4)(4,5)(5,6)(6,7)(7,8)(8,9)(9,10).. How come 72 ways?
PS:
K.. got it.. It is actually selecting 2 books adjacent and then choosing a third book= 9* 8c1.. :)
@wontmissthstime said: @AsihekAdhvaryu"Now no of ways of selecting 2 books adjacent are 9*8=72"what i think is no of ways to select 2 adjacent books is only 9 ....how comes 9*8???plz explain
Ya 9 ways you can choose 2 adjacent books but u have to choose 3 books so 3rd book u can choose from remaining 8 book So 8C1=8
total 9*8
@AsihekAdhvaryu said:Ya 9 ways you can choose 2 adjacent books but u have to choose 3 books so 3rd book u can choose from remaining 8 book So 8C1=8total 9*8
ok got it....any ideas about the question on page 37....5A 4B 3C 2D??
@KatMann said:Isn't this question to be done considering the possibility as below:1st case: Pick 5As n place any of the other three letter as 6th letter=> 1*3= 3 possibilities2nd case: Pick 4As from 5 and place the remaining two with any of the other letter=> 5c4*4c2(2 Bs from 4) + 5c4*3c2(2 Cs from 3)+ 5c4*2c2( 2 Ds) + 5c4*3(BC, CD, BD)= 65 possibilities..and so on.. this seems too lengthy.. just tell me what's the right approach
Sorry at first i overlooked D
U have to Choose 6 letter
Taking 5A so 1 letter left 1 can choose 3 ways (B,C,D)=3
Taking 4A so 2 letter left choose (BB,BC,BD,CC,CD,DD)=6
Taking 3A so 3 letter left choose (BBB,BBC,BBD,BCC,BDD,CCC,CCD,CDD)=8
Taking 2A so 4letter left choose (BBBB,BBBC,BBBD,BBCC,BBDD,BBCD,BCCC,CCCB,CCCD)=9
Taking 1A so 5letter left choose (BBBBC,BBBBD,BBBCC,BBBDD,BBBCD,BBCCC,BBCCD,BBCDD,BCCCD,BCCDD,CCCDD)=11
Taking 0A so 5letter left choose (BBBBCC,BBBBCD,BBBBDD,BBBCCC,BBBCCD,BBBCDD,BBCCCD,BBCCDD,BCCCDD)=9
TOTAL 46 BUT THERE MUST BE OTHER WAY PLEASE TELL IF ANYONE FIND ALTERNATIVE
How many times does the digit 5 appear in the numbers from 9 to 1000??
Plz help me out in solving dis problem
How many four-digit odd numbers can be formed such that every 3 in th number is follwed by a 6??
@Rshmi05 said:Plz help me out in solving dis problem
How many four-digit odd numbers can be formed such that every 3 in th number is follwed by a 6??
no three in the number --> 8*9*9*4 = 2592
3 and 6 should be mandatory pair if 3 comes and 3 cant be ending digits
so 36xy --> 1*9*4 = 36
3x6y ---> 9*4 - 36
x36y --> 8*1*4 = 32
so we have 2592 + 104= 2696
3 and 6 should be mandatory pair if 3 comes and 3 cant be ending digits
so 36xy --> 1*9*4 = 36
3x6y ---> 9*4 - 36
x36y --> 8*1*4 = 32
so we have 2592 + 104= 2696