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Let me know how to work this problem?
How many numbers are there between 500 and 1000 which have exactly one of their digist as 8?
Let me know how to work this problem?
I just figured it out. But then I had the answer in hand, not the steps (calculations) though.
if the 8 is in the 100's position then:
it can be arranged in one way
tens position can then be arranged in 9 ways (excluding 8 out of total 10 possibilities)
units position similarily in 9 ways (the other numbers can repeat)
Total: 1 x 9 x 9 = 81
if the 8 is in the 10's position then:
100s position can be arranged in 4 ways (5,6,7,9 remember the number range between 500 - 1000)
10s position in 1 way (since we assumed it to be 8)
units position in 9 ways
Total: 4 x 1 x 9 = 36
Same case if 8 is in the units position
Total: 4 x 1 x 9 = 36
Grand total of: 81+36+36 = 153 ways
One more question:
How many odd numbers less than 1000 can be formed by using the digits 0,3,5,7 when repitition of digits is not allowed?
Got one more permutation Question:
FInd the sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time.
Arun
How many odd numbers less than 1000 can be formed by using the digits 0,3,5,7 when repitition of digits is not allowed?
----------------------------------------------
Two options:
_0_
_ _ _ (no 0 at all)
(other options would either be greater than 1000 or even).
The first gives: 3x2=6 (3 options for the first digit, then 2 for the last)
The second: 3!=6 (same as building any 3-d number from 3, 5 and 7 without repetitions)
Since the options are exclusive, we simply add: 6+6=12
__________________________________________________ __________
FInd the sum of all the numbers that can be formed with the digits 2, 3, 4, 5 taken all at a time.
---------------------------------------------
There are 4! different numbers that can be formed using 2,3,4,5. Enumeration:
2345
2354
2435
2453
...
5342
5423
5432
Notice that each column contains exactly 6 occurrences of all digits, so the sum of each column is:
6x5+6x4+6x3+6x2 = 6x(5+4+3+2) = 84
If we simulate the addition we get:
__8
2345
...
+
5432
-----
___4
->
_1
_8
2345
...
+
5432
-----
__24
->
1
8
2345
...
+
5432
-----
_324
->
2345
...
+
5432
-----
93324
The answer is rite. I need to sit on Permutations and Combinations more and get a hang of it.
Arun
Assuming we're trying to find only +ve odd numbers less than 1000...Originally Posted by arunks_kumar
Single digit odd nos: 3
2-digit odd nos: 3P2 = 6
3-digit odd nos:
Without '0' --> 3! = 6
With '0' as 2nd digit --> 3P2 = 6
Total +ve odd numbers less than 1000 that satisfy the conditions = 3*6+3 = 21
How many odd numbers less than 1000 can be formed by using the digits 0,3,5,7 when repitition of digits is not allowed?
I think the ques is asking to use digits 0 3 5 7 all times..
According to this
# could be
---3 ( first 3 places filled by 0 5 or 7) = 6
---5 ( first 3 places filled by 0 3 or 7) = 6
---7 ( first 3 places filled by 0 3 or 5) = 6
total = 18
Arjmen, you are right. I forgot to count 2-digit and single-digit numbers.
You are rite Arjmen, 21 is the rite answer for the question. I too calculated the problem like Dahiya did. WTG bro!
Arun
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