6nothing mentioned abt it!!
:P
plzz just check this quesn
How many 6 digit nos. of the type a1a2a3a4a5a6 are possible having the property
a1>a2>a3<a4>a5>a6
i initially thot abt this process:
10C6 . 6C3
i.e. i am first selecting 6 nos. from 10................then selecting 3 out of those 6 and then these 3 can be arranged on either side of a4 . ( and i was gettin the rite ans as well):P
but now i think that . the 3 selected nos. a1,a2 and a3 can also be arranged b/w themselves , like its not necess. a1>a2>a3 by this method...............so what shud be the methodology i shud adopt
??
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9 Answers
1well i would go for they way they do proability sums in ncert....make a tree and then think it can be easily done
71Say a1 = 9,
So, we have Possible a2 = 8,7,6,5,4 (Hence 5 possible)
for a3 = 7,6,5,4,3 (Hence 5 possible)
for a4 = (5 possible)
Multiplying we get the total possible no when a1 = 9.
Now possible outcomes of a1 = 5
Hence Multiply that answer by 5.
This should give answer. It may be incorrect
1it indeed is..i think (cause there are more possibilities clearly!
anyways i m surprised how did akhil get right answer by his first approach.its a sheeerrrrr coincidence!...when doing elementary counting we get many expressions like [3] 1+ 1+2 + 1+2+3 + 1+2+3+4...................1+2+...8 this can be easily summed and this is just one case....more cases are there and it makes problem laborious!
6e1 i m amazed abt that kunl.......
coz otherwise it feels almost impossible to solve this
others plzz try!!
4910>a1
a1>a2
a2>a3
a3<a4
a4>a5
a5>a6
a6>a7
a7>-1
10=a1+o
a1=a2+p
a2=a3+q
a3=a4-r
a4=a5+s
a5=a6+t
a6=a7+u
a7=-1+v
__________
11=o+p+q-r+s+t+u+v
find number of solutions of the form (o,p,q,r,s,t,u,v)
That is reqd answer!! :)