1Sir, plz provide a subjective solution to the question which Subhash tried to solve at the begg. of this page
19 Answers
62The no. of 6 digit numbers which are divisible by
a) 3
b)6
a) = 900000/3
b) = 900000/6
62No. of integral solutions of
x+y+z=0
x,y,z>= -5
x>-5 means x=-5+a
y=-5+b
z=-5+c
thus,
a+b+c =15 and a, b, c are all greater than equal zero!
so we need to divide this 15 into 3 parts by using 2 dividers..
so the answer should be 17C2
62to show why i need a divisibilty test i will give a ques.
Find all 4 digit nos.divisible by 3 which constituting only 3,4,5,6,7,,9
Plz show how u can solve it without divisibility test
well abhishek in this question, you are right... it is not that you can always solve such problems the way i solved that one..
but always think this way good problems are solvable... *they will not want you to grind.. but to think :)
11Sir, i got ur point in that being 5!not 6!
to show why i need a divisibilty test i will give a ques.
Find all 4 digit nos.divisible by 3 which constituting only 3,4,5,6,7,,9
Plz show how u can solve it without divisibility test
62abhishek.. i think that for ur last question..
the only way to think is that every third number has sum of digit divisble by 3...
The other ways will be far too much more complex for comfort!
62You have 5 digit numbers... so the others wil be 3 digits only
so i think u need to find
and for the selection of rest digits for 4 remaining places we have 5C3
So the final answer should be
6C1 5C3 5!/2!
1Nishant sir,for the
The no. of 6 digit numbers which are divisible by
a) 3
b)6
question can u suugest any method using divisibility test method e.g. for 3 its that the sum of digits is divisible by 3
1Nishant sir as i have comprehended the solution could also be
no. of waya to select the two identical digits=6C1
and for the selection of rest digits for 4 remaining places we have 5C4
now the selection process is complete and now we can permute the 6 selected no.(2 identical and 4diff)by 6!/2! (as 2 are identical)
So the final answer should be
6C1 5C4 6!/2!
Plz comment
62HOW many 5 digit nos r there having 2 identical digits made from the digit s 1,2,3,4,5,6
no of ways to select 3 "single use" digits = 6C3
no of ways to select 1 "double use" digits = 3C1
No of numbers that can be made from these = 5!/2!'
I guess this is not doable..
1HOW many 5 digit nos r there having 2 identical digits made from the digit s 1,2,3,4,5,6
1selection of all possible 3 letter word(without repetation) from the word PROPORTION.
1if r=0 Σ n 1/ nCr = a and
S= 0<=i<=i<=nΣ Σ( i/nCi+ j/nCj)
Then S is =
a)na/2 c)n2a/2
b)na/4 d)na2/2