Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    2
    ReputationRep:
    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.
    Offline

    3
    ReputationRep:
    (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:
    Show
    It looks very similar to  5! = 5 \times 4 \times 3 \times 2\times 1.

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

    Once you have both, the answer should coalesce.
    Offline

    19
    ReputationRep:
    pointless question for S1
    • Thread Starter
    Offline

    2
    ReputationRep:
    (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:
    Show
    It looks very similar to  5! = 5 \times 4 \times 3 \times 2\times 1.

    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!
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.