Alevel mathematics?

Find the number of different arrangements that can be made from the 9 letters of the word JEWELLERY in which the three Es are together and the two Ls are together?

I got: 2!X3!X6! =8640

Correct answer is 6!=720

I don't understand why the answer is 6!   Can anyone help me?😇

6 Answers

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  • 1 month ago

    Rearrange the word get

    J(EEE)W(LL)RY

    If we consider those letters in the brackets

    as 1 letter, there are 6 "letters" in total.

    The possible number of arrangements of

    these 6 "letters"=6P6=6!=6*5*4*3*2*1=720.

  • 1 month ago

    I can see you are already grouping the three Es and the two Ls together.  The result is you have 6 items to arrange.

    J, (EEE), W, (LL), R, Y

    There are 6! = 720 ways to arrange the 6 items.

    But then you went a step further and said to yourself, "but the Ls can be arranged in 2! ways and the Es can be arranged in 3! ways, so we should multiply by 2! and 3! as well."

    However, the letter E is considered to be indistinguishable from the other Es. Likewise the Ls are indistinguishable from each other. So there is no need to rearrange them. There is only one arrangement of the three Es --> EEE and there is only one arrangement of the two Ls --> LL.

    That's why it is only 6! = 720.

    If you had different colored letters or they were in different fonts, for example, then it would be important to consider the order of the three Es and the two Ls. But as presented, we consider the same letters to be indistinguishable and it is not necessary to count further permutations of those letters.

  • 1 month ago

    Find the number of different arrangements that can be made from the 9 letters of the word JEWELLERY in which the three Es are together and the two Ls are together?

    (EEE) J (LL) R W Y ==> 6! = 720

  • 1 month ago

    Treat the 3 E's as one letter, and the 2 L's as one letter.  The number of permutations is

    6! = 720

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  • 1 month ago

    We need to bunch together the 3 E's as one item

    Similarly, we bunch together the 2 L's as one item

    so, EEE --> 1 item

    LL --> 1 item

    Then, J, W, R and Y are 4 separate items

    In total we are to arrange 6 items

    i.e. 6! => 720

    :)>

  • 1 month ago

    Treat the three E's as ONE letter and the two L's as ONE letter because in each case, they cannot be disassembled.  When you do this, you have only six "letters" to manipulate.  Then 6! = 720.

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