# S1 as level cie statistics help Watch

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Idk how to solve 6 i and iican someone please help , attached image

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Idk how to solve 6 i, can someone please help , attached image

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**Universecolors**)Idk how to solve 6 i, can someone please help , attached image

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Answer is 648 and 104 respectively

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#3

6ai)

The variable y can have 9 different options. therefore z now has 8 different options(1 less option since no repetition is allowed). The value of x has 9 possible combinations since its 100 - 999

6aii) for 8 _ _. the last digit must be odd so the number of combinations is 5C1. Now the middle number can be anything however 8C1 number of combinations exist to as the choice of numbers decrease to avoid repetition. Do the same type of working for the even 7 _ _ and 9 _ _ but remember the first number is already an odd number so the combination is 4C1

7 _ _

8 _ _

9 _ _

edit: typo

__X____Y____Z.__these three set of variables which are all different numbers holds all the different possible combinationsThe variable y can have 9 different options. therefore z now has 8 different options(1 less option since no repetition is allowed). The value of x has 9 possible combinations since its 100 - 999

6aii) for 8 _ _. the last digit must be odd so the number of combinations is 5C1. Now the middle number can be anything however 8C1 number of combinations exist to as the choice of numbers decrease to avoid repetition. Do the same type of working for the even 7 _ _ and 9 _ _ but remember the first number is already an odd number so the combination is 4C1

7 _ _

8 _ _

9 _ _

edit: typo

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(Original post by

6ai)

The variable y can have 9 different options. therefore z now has 8 different options(1 less option since no repetition is allowed). The value of x has 9 possible combinations since its 100 - 999

6aii) for 8 _ _. the last digit must be odd so the number of combinations is 5C1. Now the middle number can be anything however 8C1 number of combinations exist. Do the same type of working for the even 7 and 9 but remember the first number is already an odd number so the combination is 4C1

7 _ _

8 _ _

9 _ _

**zattyzatzat**)6ai)

__X____Y____Z.__these three set of variables which are all different numbers holds all the different possible combinationsThe variable y can have 9 different options. therefore z now has 8 different options(1 less option since no repetition is allowed). The value of x has 9 possible combinations since its 100 - 999

6aii) for 8 _ _. the last digit must be odd so the number of combinations is 5C1. Now the middle number can be anything however 8C1 number of combinations exist. Do the same type of working for the even 7 and 9 but remember the first number is already an odd number so the combination is 4C1

7 _ _

8 _ _

9 _ _

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#5

(Original post by

For a i , why does variable Y have 9 options , and not 8 , since one is already used in X ? Btw thanks

**Universecolors**)For a i , why does variable Y have 9 options , and not 8 , since one is already used in X ? Btw thanks

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(Original post by

6ai)

The variable y can have 9 different options. therefore z now has 8 different options(1 less option since no repetition is allowed). The value of x has 9 possible combinations since its 100 - 999

6aii) for 8 _ _. the last digit must be odd so the number of combinations is 5C1. Now the middle number can be anything however 8C1 number of combinations exist to as the choice of numbers decrease to avoid repetition. Do the same type of working for the even 7 _ _ and 9 _ _ but remember the first number is already an odd number so the combination is 4C1

7 _ _

8 _ _

9 _ _

edit: typo

**zattyzatzat**)6ai)

__X____Y____Z.__these three set of variables which are all different numbers holds all the different possible combinationsThe variable y can have 9 different options. therefore z now has 8 different options(1 less option since no repetition is allowed). The value of x has 9 possible combinations since its 100 - 999

6aii) for 8 _ _. the last digit must be odd so the number of combinations is 5C1. Now the middle number can be anything however 8C1 number of combinations exist to as the choice of numbers decrease to avoid repetition. Do the same type of working for the even 7 _ _ and 9 _ _ but remember the first number is already an odd number so the combination is 4C1

7 _ _

8 _ _

9 _ _

edit: typo

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#7

(Original post by

For 6 ii I don’t really get what u do for the middle value , can u please explain it more , im sorry : (

**Universecolors**)For 6 ii I don’t really get what u do for the middle value , can u please explain it more , im sorry : (

Also don't worry it took me some time to solve this question. I absolutely hate permutation and combination questions in statistics

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(Original post by

8 _ _. Since it must be odd the last digit must be either 1,3,5,7,9. This gives a combination of 5C1. The middle value now can accept numbers 0, 2, 4 , 6, and 4 odd numbers since one odd number is used at the end. Remember no repetition. So between 800 and 899 there are 5C1 * 8C1 = 40 different combinations.

Also don't worry it took me some time to solve this question. I absolutely hate permutation and combination questions in statistics

**zattyzatzat**)8 _ _. Since it must be odd the last digit must be either 1,3,5,7,9. This gives a combination of 5C1. The middle value now can accept numbers 0, 2, 4 , 6, and 4 odd numbers since one odd number is used at the end. Remember no repetition. So between 800 and 899 there are 5C1 * 8C1 = 40 different combinations.

Also don't worry it took me some time to solve this question. I absolutely hate permutation and combination questions in statistics

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#9

(Original post by

Thank yiuuu so muchhhh , can you please tell me how you would approach this question too 7 ii? If it’s not a bother

**Universecolors**)Thank yiuuu so muchhhh , can you please tell me how you would approach this question too 7 ii? If it’s not a bother

_ W _ _ _ _ _ _ W _

W _ _ _ _ _ _ W _ _

_ _ W _ _ _ _ _ _ W

There are 8 spaces so there are 8! different ways to arrange them however there the letter L is repeated 3 times. So for 1 arrangement of W there are 8!/3! different possible arrangements. Multiply by 3 to get the total amount of arrangements. Good luck this wednesday? (Am having my s1 exam aswell )

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the 3 different possible arrangements of W.

_ W _ _ _ _ _ _ W _

W _ _ _ _ _ _ W _ _

_ _ W _ _ _ _ _ _ W

There are 8 spaces so there are 8! different ways to arrange them however there the letter L is repeated 3 times. So for 1 arrangement of W there are 8!/3! different possible arrangements. Multiply by 3 to get the total amount of arrangements. Good luck this wednesday? (Am having my s1 exam aswell )

**zattyzatzat**)the 3 different possible arrangements of W.

_ W _ _ _ _ _ _ W _

W _ _ _ _ _ _ W _ _

_ _ W _ _ _ _ _ _ W

There are 8 spaces so there are 8! different ways to arrange them however there the letter L is repeated 3 times. So for 1 arrangement of W there are 8!/3! different possible arrangements. Multiply by 3 to get the total amount of arrangements. Good luck this wednesday? (Am having my s1 exam aswell )

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#11

(Original post by

How do u feel about the exam ? Wait u did p1 too right ? How was it for u , btw thanks for the answer I get it now

**Universecolors**)How do u feel about the exam ? Wait u did p1 too right ? How was it for u , btw thanks for the answer I get it now

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A bit scared of the permutation and combination questions and the binomials distribution questions. P1 was fine, but I realized after the exam i made a silly mistake on the trigonometry question so I lose like 5 marks there :P

**zattyzatzat**)A bit scared of the permutation and combination questions and the binomials distribution questions. P1 was fine, but I realized after the exam i made a silly mistake on the trigonometry question so I lose like 5 marks there :P

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#13

(Original post by

P1 was really bad for me , which variant were u

**Universecolors**)P1 was really bad for me , which variant were u

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(Original post by

13, wbu? and don't worry maybe you can boost the marks further with s1 or p3 next year

**zattyzatzat**)13, wbu? and don't worry maybe you can boost the marks further with s1 or p3 next year

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**zattyzatzat**)

13, wbu? and don't worry maybe you can boost the marks further with s1 or p3 next year

Can you help me with that question please

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(Original post by

i think am missing some information, can i see part 1?

**zattyzatzat**)i think am missing some information, can i see part 1?

This is i , I have already solved it

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