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How many different three-digit multiples of 5 can be composed of digits 2, 3, 4, and 5 if none of the digits is repeated?

A. 3 B. 6 C. 10 D. 12 E. 18

The units digit must be 5, otherwise the number would not be a multiple of 5. For the two other positions we have three candidates: 2, 3, and 4. Because the order of the elements matters, these two positions can be filled in \(2(C_{3}^{2}) = 6\) different ways.

Because the last digit must be 5, there is only one option, therefore, I placed a 1 in this slot. the next slot to the left, therefore, only has 3 possible options (2,3, or 4, since we've already placed the 5, and the numbers cannot repeat). The next slot only has 2 possible options. The product of these is 6.

Is there a simple way to solve this in x!/(y!z!) notation?

Because the last digit must be 5, there is only one option, therefore, I placed a 1 in this slot. the next slot to the left, therefore, only has 3 possible options (2,3, or 4, since we've already placed the 5, and the numbers cannot repeat). The next slot only has 2 possible options. The product of these is 6.

Is there a simple way to solve this in x!/(y!z!) notation?

Thanks

Your approach is correct.

As for "x!/(y!z!) notation", the solution provided uses 3C2, which is 3!/(2!1!)...
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Thanks for the clarification. I'm not familiar with the "C" notation.

nCk is choosing k items out of n distinct items so that the order of the selection does not matter and it equals to \(\frac{n!}{k!(n-k)!}\). For example, choosing 3 out of 5, 5C3, is \(\frac{n!}{k!(n-k)!}=\frac{5!}{3!(5-3)!}=10\): we can choose 10 different 3-item groups out of 5 different items.

Bunuel, I believe your annotation of your original combination may be incorrect.

You stated 2C3, whereas later you mentioned 3C2. Its a typo I believe, or I could be misinterpreting the C sign, where n is at the top and k is at the bottom.

Bunuel, I believe your annotation of your original combination may be incorrect.

You stated 2C3, whereas later you mentioned 3C2. Its a typo I believe, or I could be misinterpreting the C sign, where n is at the top and k is at the bottom.

How many different three-digit multiples of 5 can be composed of digits 2, 3, 4, and 5 if none of the digits is repeated?

A. 3 B. 6 C. 10 D. 12 E. 18

The units digit must be 5, otherwise the number would not be a multiple of 5. For the two other positions we have three candidates: 2, 3, and 4. Because the order of the elements matters, these two positions can be filled in \(2(C_{3}^{2}) = 6\) different ways.

Answer: B

I got the answer right. But why are you using combination? Since the order matters, shouldn't it be a permutation?

How many different three-digit multiples of 5 can be composed of digits 2, 3, 4, and 5 if none of the digits is repeated?

A. 3 B. 6 C. 10 D. 12 E. 18

The units digit must be 5, otherwise the number would not be a multiple of 5. For the two other positions we have three candidates: 2, 3, and 4. Because the order of the elements matters, these two positions can be filled in \(2(C_{3}^{2}) = 6\) different ways.

Answer: B

I got the answer right. But why are you using combination? Since the order matters, shouldn't it be a permutation?

We multiply 3C2 by 2, which takes care of all arrangements.
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How many different three-digit multiples of 5 can be composed of digits 2, 3, 4, and 5 if none of the digits is repeated?

A. 3 B. 6 C. 10 D. 12 E. 18

The units digit must be 5, otherwise the number would not be a multiple of 5. For the two other positions we have three candidates: 2, 3, and 4. Because the order of the elements matters, these two positions can be filled in \(2(C_{3}^{2}) = 6\) different ways.

Answer: B

I got the answer right. But why are you using combination? Since the order matters, shouldn't it be a permutation?

We multiply 3C2 by 2, which takes care of all arrangements.

Hi Bunuel, I have a follow-up question to this. If multiplying by 2 takes care of all arrangements, is it like every permutation is equal to double of combination (provided no other constraints on permutation)? Like can we just calculate combination and 2x it?

Thanks for being so kind and and answering all our queries with such patience.

Hi Bunuel, I have a follow-up question to this. If multiplying by 2 takes care of all arrangements, is it like every permutation is equal to double of combination (provided no other constraints on permutation)? Like can we just calculate combination and 2x it?

Thanks for being so kind and and answering all our queries with such patience.

No. We multiply here by 2! (2 factorial) because it's the number of arrangements of 2 distinct objects. If there were n distinct objects it would have been n!.
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