Elitmus
Exam
Numerical Ability
Permutation and Combination
How many 4 digit numbers can be formed using digits 0,1,2,3,4,5, such that number is divisible by 8??
Read Solution (Total 16)
-
- Applying the divisibility rule of 8 we get
Case 1: _ 024............(3 ways)
Case 2: _032............(3 ways)
Case 3: _104............(3 ways)
Case 4: _120............(3 ways)
Case 5: _152............(2 ways)
Case 6: _240............(3 ways)
Case 7: _304............(3 ways)
Case 8: _312............(2 ways)
Case 9: _320............(3 ways)
Case 10: _352............(2 ways)
Case 11: _432............(2 ways)
Case 12: _504............(3 ways)
Case 13: _512............(2 ways)
Case 14: _520............(3 ways)
So total combinations possible=adding above combinations we get= 3*9 + 2*5= 27+10=37 - 9 years agoHelpfull: Yes(34) No(5)
- No ending with 0 are 4120,3120,5120,1240,3240,5240,1320,4320,5320,1520,3520,4520=12 ways
no ending with 2 are 1032,4032,5032,3152,4152,4312,5312,1352,4352,1432,5432,3512,4512=13 ways
no ending with 4 are 1024,3024,5024,2104,3104,5104,1304,2304,5304,1504,2504,3504=12 ways
so total no would be 12+13+12=37 - 9 years agoHelpfull: Yes(10) No(1)
- A number is divisible by 8 if and only if the last three digit are divisible by 8
so from the following numbers cases possible
Case 1: 024 ... remaining digits 1,2,3(3 ways)
Case 2: 032.... (3 ways)
Case 3: 104...3 ways
Case 4: 120...3 ways
Case 5: 152.. As 0 cant be used so 2 ways
Case 6: 240.. 3 ways
Case 7: 304.. 3 ways
Case 8: 312.. 2 ways
Case 9: 320.. 3 ways
Case 10: 352... 2 ways
Case 11: 432.. 2 ways
Case 12: 504.. 3 ways
Case 13: 512... 2 ways
Case 14: 520.. 3 ways
Total = 37 Ans - 9 years agoHelpfull: Yes(6) No(0)
- FIRST taking no's having 0's
024
032
104
120
240
304
320
504
520
it has 9 possible way now thousand place have 3 ways therefore (9*3)=27
now taking no with non zeros
152
312
352
432
512
it has 5 possible way now thousand place have 2 ways (since 1st no cant be -ve )therefore (5*2)=10
so total possible way=27+10=37
- 9 years agoHelpfull: Yes(3) No(0)
- write ans will be 60
- 9 years agoHelpfull: Yes(2) No(4)
- the no's can be end with 0,2,4 only
=>no's end with 0 are 1240,1320,1520,3120,3240,4120,5120,5240,5320 = 9 ways
=>no's end with 2 are 1032,1352,3152,4032,4132,4312,4512,5032 = 8 ways
=> no's end with 4 are 1024,2104,3104,3024,5024,5104,5304 =7 ways
total= 9+8+7 =24 - 9 years agoHelpfull: Yes(2) No(3)
- Applying the divisibility rule of 8 we get
Case 1: _ 024............(3 ways)
Case 2: _032............(3 ways)
Case 3: _104............(3 ways)
Case 4: _120............(3 ways)
Case 5: _152............(2 ways)
Case 6: _240............(3 ways)
Case 7: _304............(3 ways)
Case 8: _312............(2 ways)
Case 9: _320............(3 ways)
Case 10: _352............(2 ways)
Case 11: _432............(2 ways)
Case 12: _504............(3 ways)
Case 13: _512............(2 ways)
Case 14: _520............(3 ways)
So total combinations possible=adding above combinations we get= 3*9 + 2*5= 27+10=37
- 9 years agoHelpfull: Yes(1) No(0)
- 3*(024)
3*(032)
3*(104)
3*(120)
2*(152)
3*(240)
3*(304)
2*(312)
3*(320)
2*(352)
2*(432)
3*(504)
2*(512)
3*(520)
-------------------------
37 ways
- 9 years agoHelpfull: Yes(0) No(0)
- Applying the divisibility rule of 8 we get
Case 1: _ 024
Case 2: _032
Case 3: _104
Case 4: _120
Case 5: _152
Case 6: _240
Case 7: _304
Case 8: _312
Case 9: _320
Case 10: _352
Case 11: _432
Case 12: _504
Case 13: _512
Case 14: _520
For all the above cases:Number of numbers formed are: 2!=2
So total combinations possible= 2 * 14=28
- 9 years agoHelpfull: Yes(0) No(2)
- Applying the divisibility rule of 8 we get
Case 1: _ 024............(3!)=6
Case 2: _032.............(3!)=6
Case 3: _104.............(3!)=6
Case 4: _120.............(3!)=6
Case 5: _152.............(2!)=2
Case 6: _240.............(3!)=6
Case 7: _304.............(3!)=6
Case 8: _312.............(2!)=2
Case 9: _320.............(3!)=6
Case 10: _352.............(2!)=2
Case 11: _432.............(2!)=2
Case 12: _504.............(3!)=6
Case 13: _512.............(2!)=2
Case 14: _520.............(3!)=6
So total combinations possible=adding above combinations we get= 6*9 + 2*5 = 54+10=64
- 9 years agoHelpfull: Yes(0) No(6)
- can I know why u are not taking 008 & 016?please reply
- 9 years agoHelpfull: Yes(0) No(1)
- 008 is nt taken just becoz no is nt repeated 00 2tyms will come if we take double zero. and 6 no will given why we take 016
- 9 years agoHelpfull: Yes(0) No(0)
- does repeation allow???thn ans wl difrnt
- 9 years agoHelpfull: Yes(0) No(0)
- 38 is correct without repetition...
Plz let me know whether 135 is the answer if rep is allowed??
Lile 5*27(cases upto hundreds digit) - 9 years agoHelpfull: Yes(0) No(1)
- here why dont we take that repeation is allowed????
can anybody plz explain..... - 9 years agoHelpfull: Yes(0) No(0)
- in question repetition information isn't given so we assume repetition
any leave that repetition
neha gupta and other guys why didn't took 016 so answer is worng - 7 years agoHelpfull: Yes(0) No(0)
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