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Maths S1 HELP!

Hello. Please can someone show me how to get this answer...

WRITE IN FACTORIAL NOTATION:

1) 8X7X6 DIVIDED BY 5X4X3

the answer in 8!x2! divided by 5!x5! according to the textbook answers but I don't know how to get it.

Cheers.
Reply 1
Original post by metaljoe
Hello. Please can someone show me how to get this answer...

WRITE IN FACTORIAL NOTATION:

1) 8X7X6 DIVIDED BY 5X4X3

the answer in 8!x2! divided by 5!x5! according to the textbook answers but I don't know how to get it.

Cheers.



You know what factorial is right?
For example, 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1

You're given 8 \times 7 \times 6

That looks very similar to 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3\times 2\times 1

But notice that we're missing the ...5 \times 4 \times 3\times 2\times 1
i.e. the rest of the terms after 6.

And notice how 5 \times 4 \times 3\times 2\times 1 = 5!

So we can argue that \dfrac{8!}{5!} = \dfrac{8 \times 7 \times 6 \times 5 \times 4 \times 3\times 2\times 1}{5 \times 4 \times 3\times 2\times 1}

That gets rid of all the terms after 6 so we're just left with \dfrac{8!}{5!} = 8 \times 7 \times 6

Now apply the same logic to 5 \times 4 \times 3

Spoiler

.

How could you get rid of all the terms after 3?

Once you have both, the answer should coalesce.
Reply 2
pointless question for S1
Reply 3
Original post by RMNDK
You know what factorial is right?
For example, 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1


You're given 8 \times 7 \times 6

That looks very similar to 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3\times 2\times 1

But notice that we're missing the ...5 \times 4 \times 3\times 2\times 1
i.e. the rest of the terms after 6.

And notice how 5 \times 4 \times 3\times 2\times 1 = 5!

So we can argue that \dfrac{8!}{5!} = \dfrac{8 \times 7 \times 6 \times 5 \times 4 \times 3\times 2\times 1}{5 \times 4 \times 3\times 2\times 1}

That gets rid of all the terms after 6 so we're just left with \dfrac{8!}{5!} = 8 \times 7 \times 6

Now apply the same logic to 5 \times 4 \times 3

Spoiler

.

How could you get rid of all the terms after 3?

Once you have both, the answer should coalesce.

Thank you very much. I know it may seem simple but I struggle with this chapter in statistics so I appreciate your help. Have a good rest of your week!

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