# 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

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

Relevance
• 2 months 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.

• 2 months 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.

• 2 months 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

• 2 months ago

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

6! = 720

• 2 months 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

:)>

• 2 months 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.