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    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:
    (a) if there are no repeated digits.

    I tried:
    Arrangements = (9P3) / (2! x 2! x 3!)
    = 21

    However, the answer is 5P3, which I understand since they are just rearranging the 5 unrepeated numbers.

    Does anyone know why my initial working was incorrect?
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    Your original working would be correct if all the numbers were different... but they’re not. If a two is chosen as the first number, a two is chosen... it doesn’t matter which one is as every two is a two. There are only five different numbers (not nine as you implied) and you’re choosing three.
    Hope this helps
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    (Original post by -Mathematician-)
    Your original working would be correct if all the numbers were different... but they’re not. If a two is chosen as the first number, a two is chosen... it doesn’t matter which one is as every two is a two. There are only five different numbers (not nine as you implied) and you’re choosing three.
    Hope this helps
    I see. But doesn’t the denominator in my working address this? Why does it work for alphabets? For example, if we were to rearrange between the six letters in “S C H O O L”, we would have to do 6!/(2!) and it would be correct.
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    https://www.examsolutions.net/tutori...=S1&topic=9999 This is one of a series of videos he has made on the topic. Give it a look as he explains it fantastically, better than I could.
 
 
 
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