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.
👍 0 👎 0 👁 23
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.
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
To account for the duplicate digits in the provided set of cards (2, 2, 3, 3, 6, 6, 6), we need to adjust the calculation.
- Out of the three digits in the number, the first slot (hundreds place) can only contain a 2 to be within the range of 200 to 300: 2 __ __
- For the second slot (tens place), we have 6 options (1, 2, 3, 3, 4, 6), considering all the available digits except for the 2 that was placed in the first slot: 2 6 _
- Finally, for the last slot (ones place), we have 5 options (1, 2, 3, 3, 4), considering all the available digits excluding the ones used in the previous slots: 2 6 5
To find the total number of different numbers that can be formed in this way, we multiply the number of options for each slot: 1 option for the first slot, 6 options for the second slot, and 5 options for the third slot.
1 x 6 x 5 = 30
Therefore, there are 30 different numbers that can be formed with the provided cards, not accounting for the range limitation.
However, since we are only considering numbers between 200 and 300, we need to subtract the numbers that are outside of this range.
The numbers that fall outside of the range are: 265, 253, 264, 263, 213, 246, 243, 236, 226, 233, 232.
Therefore, the total number of different numbers between 200 and 300 that can be formed with the provided cards is 30 - 11 = 19.
To further clarify, the reason we need to adjust for duplicate digits is that there are two 2's and three 6's among the nine cards. When we choose one of the 2's for the first slot, we are not able to distinguish it from the other 2 because the cards are identical. Similarly, when we choose one of the 6's for the second or third slot, we cannot distinguish it from the other 6's.
To adjust for this, we divide the total number of possible arrangements (2 x 8 x 7) by the number of indistinguishable arrangements for each duplicate digit. In this case, there are 2 duplicate digits (2 and 6), so we divide by 2! (the factorial of 2) for the 2's and 3! (the factorial of 3) for the 6's.
Therefore, the total number of different numbers that can be made between 200 and 300 using three cards from the given nine cards is:
(2 x 8 x 7) / (2! x 3!) = 2 x 4 x 7 = 56
However, we need to consider that only the numbers between 200 and 300 are valid. So, we need to subtract the numbers that are less than 200 or greater than 300.
Out of the 56 possible combinations, we subtract the 2 possible combinations with a 1 in the first slot (1 cannot be used as it would make the number less than 200) and the 10 possible combinations with a 4 or 6 in the first slot (4 or 6 would make the number greater than 300).
Therefore, the final answer is: 56 - 2 - 10 = 44.
So, there are 44 different numbers that can be made between 200 and 300 using three cards from the given set.