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Numerical Ability
Permutation and Combination
a person starts writting four digit numbers, how many times did he write 2
Read Solution (Total 12)
-
- total 4 digit no. = 9*10*10*10 = 9000
not containing 2 = 8*9*9*9 = 5832
total 4 digit number contain 2 = 9000 - 5832 = 3168
So ans:3168 - 8 years agoHelpfull: Yes(23) No(5)
- lets see from 0-9 --> 1
from 10-19 --> 1
from 20-29 --> 11
from 30-39 --> 1
from 40-49 --> 1
from 50-59 --> 1
from 60-69 --> 1
from 70-79 --> 1
from 80-89 --> 1
from 90-99 --> 1
which makes 20 from 0 to 99
now please observer that similar pattern follows for
000 - 099 -->20
100 - 199 --> 20
300 - 399 --> 20
400 - 499 --> 20
500 - 599 --> 20
600 - 699 --> 20
700 - 799 --> 20
800 - 899 --> 20
900 - 999 --> 20
however for 200 to 299 we have that extra 2 as first digit for all these 100 numbers, so
200 - 299 --> 120
so
000 - 999 -->300
so for
1000 - 1999 -->300
3000 - 3999 -->300
4000 - 4999 -->300
5000 - 5999 -->300
6000 - 6999 -->300
7000 - 7999 -->300
8000 - 8999 -->300
9000 - 9999 -->300
again for
2000 - 2999 -->300 + 1000 as we have that extra 2 in the first digit for all these numbers, hence
2000 - 2999 -->1300
So total number of times digit 2 is used 1000 - 9999 is 3700
Join The Disc - 8 years agoHelpfull: Yes(16) No(0)
- Let the four digit number be abcd.
Case 1: 2 appears only once in the 4-digit number.
No. of ways of having a number of the form -
2bcd = 9*9*9 = 729
a2cd = 8*9*9 = 648
ab2d = 8*9*9 = 648
abc2 = 8*9*9 = 648
Total no. of ways = 2673
Total no. of times 2 appears = 2673
Case 2: 2 appears only twice in the 4-digit number.
No. of ways of having a number of the form -
22cd = 9*9 = 81
2b2d = 9*9 = 81
2bc2 = 9*9 = 81
a2c2 = 8*9 = 72
ab22 = 8*9 = 72
a22d = 8*9 = 72
Total no. of ways = 459
Total no. of times 2 appears = 459*2 = 918
Case 3: 2 appears only thrice in the 4-digit number.
No. of ways of having a number of the form -
222d = 9
22c2 = 9
2b22 = 9
a222 = 8
Total no. of ways = 35
Total no. of times 2 appears = 35*3 = 105
Case 4: 2 appears four times in the 4-digit number.
2222 i.e. only once
Total no. of times 2 appears = 4
So, total no. of times 2 appears = 2673+918+105+4 = 3700. - 8 years agoHelpfull: Yes(9) No(0)
- Ans:4000
X X X X
for thousand's position no of 2's 1×1000=1000
for hundredth position no of 2's
10×100=1000
for tenth's position no of 2's
10×10×10=1000
for one's position no of 2's
1000
So ans:4000 - 8 years agoHelpfull: Yes(3) No(5)
- use this formula :n*10^(n-1)
where n is equal to no of digits u want.. - 8 years agoHelpfull: Yes(3) No(0)
- 25 common sence
- 8 years agoHelpfull: Yes(2) No(7)
- When 2 is at unit place=9*10*10*1
When 2 is at tenth place=9*10*1*10
When 2 is at hundred place=9*1*10*10
When 2 is at thousand place = 1*10*10*10
So Total no. = 900 + 900 + 900 + 1000 = 3700 - 6 years agoHelpfull: Yes(2) No(0)
- 0 as one girl is already there
- 7 years agoHelpfull: Yes(1) No(1)
- 1 digit---- 1 time
10---------- (1*10)+10=20
100---------(20*10)+100=300
1000--------(300*10)+1000=4000 - 8 years agoHelpfull: Yes(0) No(4)
- total no. ways a 4 digit no can be formed=9*10*10*10
total no ways in which a 4 digit no can be formed without using 2=8*9*9*9
total no ways in which a 4 digit no can be formed using 2=(9*10*10*10)-(8*9*9*9)=3168 - 7 years agoHelpfull: Yes(0) No(0)
- i don't know
- 6 years agoHelpfull: Yes(0) No(0)
- When 2 is at unit place=9*10*10*1
When 2 is at tenth place=9*10*1*10
When 2 is at hundred place=9*1*10*10
When 2 is at thousand place = 1*10*10*10
So Total no. = 900 + 900 + 900 + 1000 = 3700 - 4 years agoHelpfull: Yes(0) No(0)
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