Elitmus
Exam
Numerical Ability
Permutation and Combination
how many number of numbers can by formed by using the digits 1,2,3,4 and 5 which are divisible by 6?
Read Solution (Total 19)
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- 3 digit numbers formed using 1,2,3,4,5 divisible by 6
unit digit should be 2/4
no. can be XY2 & XY4
X+Y+2 = 6,9 & X+Y+4 = 9,12
X+Y = 4,7 & X+Y = 5,8
(X,Y)= (1,3),(3,1),(2,5),(5,2) &
(X,Y)= (2,3),(3,2),(3,5),(5,3)
total 8 numbers without repetition. - 10 years agoHelpfull: Yes(46) No(21)
- For 1 digit=0
for 2 digit=4
for 3 digit=8
for 4 digit=12
for 5 digit=48
total=72
- 10 years agoHelpfull: Yes(36) No(1)
- divisible by 6 means ->should be divisible by 2 and 3.
1) for 2 , the unit place should have even number ,thus 4!=24
2) for 3, the sum is 18 means divisible by 3 ,thus 4!=24
thus 24+24=48.
- 10 years agoHelpfull: Yes(9) No(5)
- HOW MANY 3 DIGIT NUMBERS
- 10 years agoHelpfull: Yes(4) No(1)
first of all
single digit no . =0
2 digit no(12,24,42,54)=4
3 digit no=12
4digit no=24
5digit no=24
totalno=64- 10 years agoHelpfull: Yes(4) No(6)
- Qn : How many three digit numbers can be formed (without repetition) by using the digits 1,2,3,4 and 5 which are divisible by 6?
Ans : For 6, both 2 & 3 division rules should be satisfied.
So Unit digit should be 2 or 4
Numbers can be XY2 & XY4
X+Y+2 & X+Y+4 should be divisible by 3.
(X,Y,2)= (1,3,2),(3,1,2),(3,4,2),(4,3,2)
(X,Y,4)= (2,3,4),(3,2,4),(3,5,4),(5,3,4)
So 8 numbers are possible. - 8 years agoHelpfull: Yes(4) No(0)
- rakesh how is it possible...u told that with out repetition..bt u consider,( 5,2).there is already 2 as a no...it should be ...4,3
- 10 years agoHelpfull: Yes(3) No(0)
- for 6 number divisible by 2 and 3
count =0
_12 only 3 count = 1
_14 not possible
_24 or _42 only 3 count =3
_32 ...2 and 1 count = 5
_34....5 and 4 count = 7
_52 not possible
_54 only 3 count =8
if number not repeated
ans is 8 - 10 years agoHelpfull: Yes(2) No(3)
- not complete question
- 10 years agoHelpfull: Yes(2) No(1)
- rakesh,,how u took x+y+2=9 and x+y+4=9 ,when the sum should be divisible by 2 ,3
explain
plzzz
- 10 years agoHelpfull: Yes(1) No(0)
- 12,42,24,54,132,312,342,432,324,234 total 10 nos
- 10 years agoHelpfull: Yes(0) No(2)
- 152,132,312,342,432,512,234,324,354,534,
total 10 no
- 10 years agoHelpfull: Yes(0) No(4)
- (11)
if repitition allowed
112,152,132,312,342,432,512,234,324,354,534 - 10 years agoHelpfull: Yes(0) No(1)
- ______ _____ _______ _____1___ ______2____ -------->1X2x3x1x1----->6 ways
_______ _____ _______ _____2____ ______4___ ---------->6 ways
______ ______ ______ ______5____ _____4____ -------->6 ways
ans----->18 - 10 years agoHelpfull: Yes(0) No(0)
- @RAKESH how can you assume three digits no.
if we assume two digits no then three more no which is disible by 6
similarly for four digits no and 5 digits no as well. - 10 years agoHelpfull: Yes(0) No(0)
- NOTE: A number is divisible by 6 if it is even and if the sum of its digits is divisible by 3.
NOW in this case sum of all the digits (1+2+3+4+5=15) is divisible by 3.
Now only 2 option arise to get an even number.
case 1. __ __ __ __ 2 or
case 2. __ __ __ __ 4
so the required combination will be .... for case 1.- 4*3*2*1=24 Case 2- 4*3*2*1= 24
hence 24+24=48 is the answer - 9 years agoHelpfull: Yes(0) No(0)
- one digit number 0
two digit number (12)(24)(42)
three digit number (132)(312)(354)(534)
four digit number (1254)(1524)(5124)(5214)(2154)(2514)(1452)(1542)(5412)(5142)(5142)(4512)
five digit number (ending with 4 is 4*3*2*1= 24)(ending with 2 is 4*3*2*1=24)
total= 67 - 9 years agoHelpfull: Yes(0) No(1)
- Nimber of Numbers to be formed, i.e
1 digit no. - 0
2 digit no. - 4 --> (12,24,42,54)
3 digit no. - 8 --> combination of possible numbers (234) and (345)
4 digit no. - 24 --> combination of possible numbers (1245)
and
5 digit no. - 48 --> combination of possible numbers (12345)
so total possible numbers are 0 + 4 + 8 + 24 + 48 = 84 - 7 years agoHelpfull: Yes(0) No(0)
- 1 digit number=0
2 digit number=12,24,42,54 =>total=4
3 digit number=132,312,432,342,234,324,534,354=>total=8
4 digit number:
last digit of each number must be 4 or 2 so,
for last digit as 4: 1254 the 1st 3 digits arrange them self in 3! ways so total number=6
for last digit as 2:1452 the 1st 3 digits arrange them self in 3! ways so total number=6
hence total 4 digit number=6+6=12
now 5 digit number=4! x 2!=48
so total numbers form=>4+8+12+48=72(ans) - 7 years agoHelpfull: Yes(0) No(0)
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