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umaa
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 Posted: Tue Jul 08, 2008 10:32 pm Post subject: permutation, combination |
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How many 5 digit numbers can be formed which are divisible by 3 using the numerals 0,1,2,3,4,5 (WITHOUT REPETITION)
a. 216
b. 3152
c. 240
d. 600
The answer is 216. |
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Ian Stewart
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| Posted: Wed Jul 09, 2008 5:48 am Post subject: Re: permutation, combination |
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| umaa wrote: | How many 5 digit numbers can be formed which are divisible by 3 using the numerals 0,1,2,3,4,5 (WITHOUT REPETITION)
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If a number is divisible by 3, the sum of its digits is divisible by 3. We need to choose five digits from {0,1,2,3,4,5} which add to a multiple of 3. We could choose:
{1, 2, 3, 4, 5}
or
{0, 1, 2, 4, 5}
If we use the digits {1, 2, 3, 4, 5}, we have 5 choices for the first digit, 4 for the second, etc- 5! = 120 numbers we can make in total.
If we use the digits {0, 1, 2, 4, 5}, we only have 4 choices for the first digit (because it cannot be zero), 4 choices for the second digit, 3 for the third, etc- 4*4! = 96 numbers we can make in total.
120+96 = 216.
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umaa
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| Posted: Wed Jul 09, 2008 6:04 am Post subject: |
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preetha_85
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| Posted: Wed Jul 09, 2008 6:35 am Post subject: |
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Hi ,
for a no. to be divisible by 3 the nos should add up to a multiple of 3.
Given the digits (0,1,2,3,4,5)
Only the combinations given below will add up to a multiple of 3 :
1. (1,2,3,4,5)
2. (0,1,2,4,5)
Case 1:
these nos can be arranged in 5! ways i.e 120
Case 2:
Since if 0 is the 1st digit it will not form a 5 digit no. they can be arranged in :4*4*3*2*1
hence the total permutations is :120 +96 =216 |
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preetha_85
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Joined: 06 Jul 2008
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Test Date: Aug 23
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| Posted: Wed Jul 09, 2008 6:35 am Post subject: |
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Hi ,
for a no. to be divisible by 3 the nos should add up to a multiple of 3.
Given the digits (0,1,2,3,4,5)
Only the combinations given below will add up to a multiple of 3 :
1. (1,2,3,4,5)
2. (0,1,2,4,5)
Case 1:
these nos can be arranged in 5! ways i.e 120
Case 2:
Since if 0 is the 1st digit it will not form a 5 digit no. they can be arranged in :4*4*3*2*1
hence the total permutations is :120 +96 =216 |
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evansbd
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| Posted: Thu Jul 17, 2008 8:47 am Post subject: Interesting |
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I noticed most people solved this problem exactly the same way.
I did the last part a little different.
Case 1: 5! = 120 - this is the easy part
Case 2:
I kind of understand why people will do 4*4!. However if I were on the exam I probably would not have realized I needed to multiply by four.
Personally, I saw that 0 couldn't be the first number for a 0 1 2 4 5 combination. I could see that I needed to find the number of combinations then subtract the ones that didnt fit the criteria of being divisible by 3. So I said there are 5! combinations of 0 1 2 4 5. However we must subtract the number of combinations where 0 is the first number.
total combinations - 0 first combinations = 5! - 4! = 120 - 24 = 96
Then I added 120 + 96 = 216 |
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