permutations

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carebear

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Using the digits 2,2,2,3,3,4,5:
How many are greater than 3,400,000 and divisible by 5?

Case 1:

first is 3, second is 4, last is 5: (1 x 1 x 4 x 3 x 2 x 1x 1) / (3!2!) = 2 ways

Case 2:

first is 4, last is 5: (1 x 5 x 4 x 3 x 2 x 1 x 1)/(3!2!) = 10 ways

2 + 10 =12 ways but the answer says 14 ways...please help...what am I missing?
 
Since there are two 3's to choose from, and we can form 34 or 43, there are 4 ways to arrange the beginning two numbers.

Since the 5 is the only choice for the end, we can arrange the 4 center numbers: 2,2,2,3 in 4 ways.

4*4=16

But, we have to subtract off the two numbers we repeated because of the two 3's.

16-2=14.
 
carebear said:
Using the digits 2,2,2,3,3,4,5:
How many are greater than 3,400,000 and divisible by 5?

Case 1:

first is 3, second is 4, last is 5: (1 x 1 x 4 x 3 x 2 x 1x 1) / (3!2!) = 2 ways

Case 2:

first is 4, last is 5: (1 x 5 x 4 x 3 x 2 x 1 x 1)/(3!2!) = 10 ways

2 + 10 =12 ways but the answer says 14 ways...please help...what am I missing?
Click to expand...

For case 1 - you'll fill 4 positions with one 3 and three 2. We can put the 3 uniquely in any of those four positions - so there are four ways.
 
Hello, carebear!

Using the digits {2, 2, 2, 3, 3, 4, 5},
How many are greater than 3,400,000 and divisible by 5?
Click to expand...

To be divisible by 5, the last (rightmost) digit must be 5.

There are only two cases to consider:
. . (1) the first digit is 3
. . (2) the first digit is 4


(1) If the first digit is 3, the second digit must be 4.

. . .We have: .\(\displaystyle 3\:4\:\_\:\_\:\_\:\_\;5\)
. . .\(\displaystyle \text{The other digits }\{2,2,2,3\}\text{ can be arranged in }\frac{4!}{3!} \:=\: 4\text{ ways.}\)


(2) The first digit is 4.

. . .We have: .\(\displaystyle 4\:\_\:\_\:\_\:\_\:\_\:5\)

. . \(\displaystyle \text{The other digits }\{2,2,2,3,3\}\text{ can be arranged in }\frac{5!}{3!\,2!} \:=\:10\text{ ways.}\)


Therefore, there are: .\(\displaystyle 4 + 10 \:=\:14\) such numbers.

 
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