Nine cards are numbered 1, 2, 2, 3, 3, 4, 6, 6, 6.Three of the nine cards are chosen and placed in a line, making a 3-digit number. Find how
many different numbers can be made in this way if the number is between 200 and 300.
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asked by Raj
today at 12:27pm
There are three digits in your number ___ ___ ___
We sometimes refer to them as "slots"
The first slot is where the hundreds is placed.
For your number to be between 200 and 300 only the 2's could go in that slot
_ 2_ , ___, ___
Then there are 8 other numbers that could go in the next slot (since one of the two 2's is used up in the first slot), then 7 numbers left for the next slot
2 x8 x 7
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👩🏫
Ms Pi 3.14159265358979323
today at 12:53pm
be sure to adjust for duplicate digits, since the equal digits are indistinguishable.
The answer should be 22.
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asked by Raj
today at 2:14pm
Brute force gets you the answer of 22 : )
I am working on a formula method right now.
But here is brute force : )
212, 213, 214, 216, 221, 223, 224, 226, 232, 233, 231, 234, 236, 241, 242, 243, 246, 261, 262, 263, 266, 264
Can you tell me formula method?
It really doesn't matter that there are multiples of 6, 3, etc until we get to the cases
the only numbers possible for picking are 1,2,3,4,6
To have a 3 digit number that lies between 200 and 300 implies that the
number must take the form 2qq
There are doubles for 2,3, and 6
case 1: 2qq, where qq is one of the doubles
2 such cases , namely 233, 266
case 2: 2qq, where the qq is made up of 2 different ones
(1)(5 possible #'s)(4 possible #'s) <---- can't have two the same, that was case1
= 1()(5)(4) = 20
total number of cases = case1+ case2 = 2+20 = 22
there is no formula that will get you the 22 directly.
To find the number of different numbers that can be made between 200 and 300 using the given cards, you can use the following formula:
1. Identify the restrictions: The first digit must be 2 and there can be no repeated digits.
2. Create a list of the available digits: In this case, the available digits are 1, 2, 3, 4, and 6 (excluding the duplicates).
3. Determine the number of choices for each digit slot:
- For the first digit slot, there is only 1 choice, which is 2.
- For the second digit slot, there are 4 choices remaining (1, 3, 4, and 6).
- For the third digit slot, there are 3 choices remaining (1, 3, and 4).
4. Multiply the number of choices for each slot: 1 x 4 x 3 = 12.
- This gives you the total number of permutations of the available digits for the three-digit number.
5. Adjust for indistinguishable duplicates: Since there are two 2's and two 3's, you need to divide the total permutations by the number of duplicate permutations.
- The number of duplicate permutations of the 2's is 2! (2 factorial) = 2.
- The number of duplicate permutations of the 3's is also 2! = 2.
- Divide the total permutations by the product of the duplicate permutations: 12 / (2 x 2) = 3.
6. The final answer is 3. There are 3 different numbers that can be made between 200 and 300 using the given cards.
To summarize:
- Number of choices for the first digit slot: 1
- Number of choices for the second digit slot: 4
- Number of choices for the third digit slot: 3
- Total permutations: 1 x 4 x 3 = 12
- Adjusted for duplicates: 12 / (2 x 2) = 3
So, there are 3 different numbers that can be made in this way if the number is between 200 and 300.
Sure! To solve this problem using a formula method, we can break it down into several steps:
Step 1: Count the total number of ways to choose 3 cards from the given set of 9 cards. This can be calculated using the combination formula:
C(n, r) = n! / (r!(n-r)!),
where n is the total number of cards and r is the number of cards to be chosen. In this case, n = 9 and r = 3. So we have:
C(9, 3) = 9! / (3!(9-3)!) = 9! / (3!6!) = (9 * 8 * 7) / (3 * 2 * 1) = 84.
Step 2: Count the number of ways to choose 3 cards such that the number formed is between 200 and 300.
Since the hundreds digit can only be 2, we have already accounted for that. So let's focus on the tens and units digits.
- If the tens digit is 2, then the units digit can only be 1 or 3. So there are 2 possibilities.
- If the tens digit is not 2, then there are 8 possibilities for the tens digit (excluding the two 2's). And since the units digit cannot be the same as the tens digit, there are 7 possibilities for the units digit.
So the total number of ways to choose 3 cards to form a number between 200 and 300 is:
2 + (8 * 7) = 2 + 56 = 58.
Therefore, there are 58 different numbers that can be made in this way.