@ravi.theja said:zero??
should be zero no doubt.
hw many 6 digit no contain exactly 4 diff digits?
@vbhvgupta said:Can you solve it?
a+(n-1)d=1/m=>ma+(mn-m)d=1
a+(m-1)d=1/n=>na+(mn-n)d=1
equating both, we get.
a(m-n)=(m-n)d or a=d
thus, a+(n-1)a=1/m
=>a(1+n-1)=1/m or a=d=1/mn
sum to mn terms,
S=mn/2[2/mn+(mn-1)/mn]=mn/2[1/mn + 1]=(mn+1)/2
@mailtoankit said:(mn+1)/2 ?a+(m-1)d=1/na+(n-1)d=1/msolving...a=d=1/mnmn/2(2/mn+(mn-1)/mn)=(mn+1)/2
(mn+1)/2
Find the 2nd term of an AP if the sum if its first 5 even terms is 15 and the sum of the first 3 terms equal to -3.
@The_Loser said:hw many 6 digit no contain exactly 4 diff digits?
the number can be either of the form aaabcd or aabbcd
casei) aaabcd
4 nos can be selected in 10C4 ways. no to be repeated in 4C1 ways.internal arrangements possible in aaabcd=6!/3!
also 0 shouldn't be the first digit
thus, no of ways=10C4*4C1*6!/3!*9/10
caseii) aabbcd
4 nos can be selected in 10C4 ways. nos to be repeated in 4C2 ways. internal arrangements possible=6!/2!*2!
also 0 shouldn't be the first digit
thus, no of ways=10C4*4C2*6!/2!2!*9/10
add these. this must be the solution.
EDIT: comes out to be 294840
@vbhvgupta said:Find the 2nd term of an AP if the sum if its first 5 even terms is 15 and the sum of the first 3 terms equal to -3.
a+5d=3
a+d=-1
d=1 & a=-2
thus, -1 ?
a+d=-1
d=1 & a=-2
thus, -1 ?
@The_Loser said:hw many 6 digit no contain exactly 4 diff digits?
No zeroes
9c4*(6!/2!2! +6!/3!) =126(180+60)=30240
Containing zeroes
9c3(5*5!/2!2! +5*5!/3!)=84(250)=21000
51240??
@vbhvgupta said:Find the 2nd term of an AP if the sum if its first 5 even terms is 15 and the sum of the first 3 terms equal to -3.
sum of the first 3 terms=>a+a+d+a+2d=-3
=>3a+3d=-3=>a+d=-1
2nd term is -1
@vbhvgupta said:Find the 2nd term of an AP if the sum if its first 5 even terms is 15 and the sum of the first 3 terms equal to -3.
5/2(2a + 4d) = 15
2a+4d=6
and 3/2(2a+2d)=-3
a+d=-1
2(-1-d)+4d=6
-2+2d=6
d=4 so a=-5
hence 2nd term = -5+4 = -1
@x2maverickc said:the number can be either of the form aaabcd or aabbcdcasei) aaabcd4 nos can be selected in 10C4 ways. no to be repeated in 4C1 ways.internal arrangements possible in aaabcd=6!/3!also 0 shouldn't be the first digitthus, no of ways=10C4*4C1*6!/3!*9/10caseii) aabbcd4 nos can be selected in 10C4 ways. nos to be repeated in 4C2 ways. internal arrangements possible=6!/2!*2!also 0 shouldn't be the first digitthus, no of ways=10C4*4C2*6!/2!2!*9/10add these. this must be the solution.EDIT: comes out to be 294840
perfect
hw many no's smaller dan 2*10^8 & are divisible by 3 can be written by means f digits 0,1 & 2. (exclude single & double digit numbers).
@Tiws
@Tiws said:Let X = {a, b, c} and Y = {l, m}. Consider the following four subsets of X Ä‚— Y. F1 = {(a, l), (a, m), (b, l), (c, m)} F2 = {(a, l), (b, l), (c, l)} F3 = {(a, l), (b, m), (c, m)} F4 = {(a, l), (b, m)} Which one, amongst the choices given below, is a representation of functions from X to Y? (A) F1, F2 and F3 (B) F2, F3 and F4 (C) F2 and F3 (D) F3 and F4 (E) None of the above
c) f2 and f3
f2 & f3 being into functions
@The_Loser said:hw many no's smaller dan 2*10^8 & are divisible by 3 can be written by means f digits 0,1 & 2. (exclude single & double digit numbers).
not sure though but worth a try.
we see the all the nos that'll appear will be in base 3.
so total nos
subtract single and double digits
ans should be 2*3^8 - 9
Q3
@vbhvgupta said:Q3
b)21578
1000(1+5/100)^14+1000(1+5/100)^13+...............1000(1+5/100)^0
=1000*((1.05)^15-1)/0.05
1000(1+5/100)^14+1000(1+5/100)^13+...............1000(1+5/100)^0
=1000*((1.05)^15-1)/0.05
@The_Loser said:hw many no's smaller dan 2*10^8 & are divisible by 3 can be written by means f digits 0,1 & 2. (exclude single & double digit numbers).
2*10^8 is 200000000.
it is a 9 digit number.
total no. of numbers=3^9
to subtract: 2xxxxxxxx where at least one x is 1 or 2.
it is 3^8
For single/two digit numbers: 0,1,2,10,11,12,20,21,22:9
ans=3^9-3^8-9
it is a 9 digit number.
total no. of numbers=3^9
to subtract: 2xxxxxxxx where at least one x is 1 or 2.
it is 3^8
For single/two digit numbers: 0,1,2,10,11,12,20,21,22:9
ans=3^9-3^8-9