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# Probability

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Using the digit 2,3,4,5,6 without repition. How many

a) numbers greater than 40000 can be formed?

b) even numbers greater than 40000 can be formed?

c) odd numbers greater than 40000 can be formed?

Oct 1, 2017

#1
+97500
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Using the digit 2,3,4,5,6 without repition. How many

a) numbers greater than 40000 can be formed?

the first digit can be 4, 5, or 6 and the others can be any.  So that is

3*4*3*2*1  = 72 ways

b) even numbers greater than 40000 can be formed?

The last can be 2 or 4 or 6 and the first can be 4 or 5 or 6

If the last on is 4 or 6 then there is    2*2*3*2*1 = 24

If the last digit is 2 then there is         1*3*3*2*1  = 18

You have to add those two result together to get the total number.   42 ways

c) odd numbers greater than 40000 can be formed?

The last can be 3 or 5 and the first can be 4 or 5 or 6

If the last on is 3 then there is            1*3*3*2*1  =  18

If the last digit is 5 then there is         1*2*3*2*1  =   12

You have to add those two result together to get the total number.     30 ways

Oct 1, 2017
edited by Melody  Oct 1, 2017
edited by Melody  Oct 1, 2017
#2
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LOL!!!!!

Doesn't every poster have to accept our answers at their own peril ????

Oct 1, 2017
#4
+97500
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Yes absolutely, although there is a high probability that probablility questions isare more perilous than most other types of questions.   LOL

Melody  Oct 2, 2017
#3
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When the number begins with 4, then there are 4! =24 permutations for 2, 3, 5, 6

Half of them, or 12 would be even, and the other half or 12, odd.

When the number begins with 5, then there are 4! =24 permutations for 2, 3, 4, 6

3/4  x  24 =18 would be even, and 1/4 x 24 =6 would be odd.

When the number begins with 6, then there are 4! = 24 permutations for 2, 3, 4, 5

Half of them or 12 would be even, and the other half or 12 would be odd.

So, 24+24+24 =72 numbers over 40,000, and:

12+18+12 =42 would be even, and:

12 + 6+ 12=30 would be odd.

Just as Melody found !!

Oct 1, 2017
edited by Guest  Oct 1, 2017