I never know how to do these questions and they're usually what I lose most marks on. Can someone help me with how to do this one?

Original post by Adidas02

I never know how to do these questions and they're usually what I lose most marks on. Can someone help me with how to do this one?

Start with a full deck. How many different possibilities are there for Shelley to receive?

Now you have a full deck minus 1 card you gave away.

Let's say the cards are labelled Card 1, Card 2, ..., Card 25.

If Shelley receives Card 1, how many different possibilities are there for Pauline to receive?

If Shelley receives Card 2 instead, how many cards for Pauline to choose from again?

[...]

If Shelley receives Card 25 instead, how many cards for Pauline to choose from again?

Then adding up these numbers will give the total number of possibilities, which can be written down straight away, can you see what it is?

Original post by Adidas02

I never know how to do these questions and they're usually what I lose most marks on. Can someone help me with how to do this one?

How many choices are there for the card given to Shelley?

How many choices are there for the card given to Pauline? (Remember that a card has already been given to Shelley).

Then use the "product rule for counting" to multiply these numbers to get the final answer.

Original post by RDKGames

Start with a full deck. How many different possibilities are there for Shelley to receive?

Now you have a full deck minus 1 card you gave away.

Let's say the cards are labelled Card 1, Card 2, ..., Card 25.

If Shelley receives Card 1, how many different possibilities are there for Pauline to receive?

If Shelley receives Card 2 instead, how many cards for Pauline to choose from again?

[...]

If Shelley receives Card 25 instead, how many cards for Pauline to choose from again?

Then adding up these numbers will give the total number of possibilities, which can be written down straight away, can you see what it is?

Now you have a full deck minus 1 card you gave away.

Let's say the cards are labelled Card 1, Card 2, ..., Card 25.

If Shelley receives Card 1, how many different possibilities are there for Pauline to receive?

If Shelley receives Card 2 instead, how many cards for Pauline to choose from again?

[...]

If Shelley receives Card 25 instead, how many cards for Pauline to choose from again?

Then adding up these numbers will give the total number of possibilities, which can be written down straight away, can you see what it is?

Thanks So, 600 (24 × 25)?

How about the second question?

um how would i work out the 2nd one, ever since i saw a vid on ! ive struggled with those questions and over complicated

Original post by Adidas02

Thanks So, 600 (24 × 25)?

How about the second question?

How about the second question?

Consider the lists individually, then add their totals together.

For the first list order, if he chooses 1 boy (how many possibilities?) then he can still choose from the same amount of girls. Once a girl is chosen (how many possibilities?), he can only choose between 11 boys. Then multiply these 3 number of possibilities together for the total amount for List order of boy, girl, boy.

Repeat similarly for the second list order.

(edited 6 years ago)

I'M NOT 100% SURE THIS IS RIGHT, IT'SJUST HOW I'D WORK IT OUT

For [Boy, Girl, Boy] There are 10 boys, so theres a 1/10 chance he'll pick a boy, Then there are 12 girls, so 10×12=120 different lists so far. (because if he picks boy 1, and then there are 12 girls, if he picks boy 2 there are still 12 girls). for the last bit there are 9 boys left bc one of them have already been picked. so 120 x 9 = 1080. 1080 possibilities. For [Girl, Boy Girl] its a similar process but aince there are 12 girls it would be 12x10= 120. then 11 girls left so 120x11

For [Boy, Girl, Boy] There are 10 boys, so theres a 1/10 chance he'll pick a boy, Then there are 12 girls, so 10×12=120 different lists so far. (because if he picks boy 1, and then there are 12 girls, if he picks boy 2 there are still 12 girls). for the last bit there are 9 boys left bc one of them have already been picked. so 120 x 9 = 1080. 1080 possibilities. For [Girl, Boy Girl] its a similar process but aince there are 12 girls it would be 12x10= 120. then 11 girls left so 120x11

Original post by Rojin001

I'M NOT 100% SURE THIS IS RIGHT, IT'SJUST HOW I'D WORK IT OUT

For [Boy, Girl, Boy] There are 10 boys, so theres a 1/10 chance he'll pick a boy, Then there are 12 girls, so 10×12=120 different lists so far. (because if he picks boy 1, and then there are 12 girls, if he picks boy 2 there are still 12 girls). for the last bit there are 9 boys left bc one of them have already been picked. so 120 x 9 = 1080. 1080 possibilities. For [Girl, Boy Girl] its a similar process but aince there are 12 girls it would be 12x10= 120. then 11 girls left so 120x11

For [Boy, Girl, Boy] There are 10 boys, so theres a 1/10 chance he'll pick a boy, Then there are 12 girls, so 10×12=120 different lists so far. (because if he picks boy 1, and then there are 12 girls, if he picks boy 2 there are still 12 girls). for the last bit there are 9 boys left bc one of them have already been picked. so 120 x 9 = 1080. 1080 possibilities. For [Girl, Boy Girl] its a similar process but aince there are 12 girls it would be 12x10= 120. then 11 girls left so 120x11

I haven't checked your working but please don't post full solutions - post hints instead.

Please read the posting guidelines.

Original post by RDKGames

Consider the lists individually, then add their totals together.

For the first list order, if he chooses 1 boy (how many possibilities?) then he can still choose from the same amount of girls. Once a girl is chosen (how many possibilities?), he can only choose between 11 boys. Then multiply these 3 number of possibilities together for the total amount for List order of boy, girl, boy.

Repeat similarly for the second list order.

For the first list order, if he chooses 1 boy (how many possibilities?) then he can still choose from the same amount of girls. Once a girl is chosen (how many possibilities?), he can only choose between 11 boys. Then multiply these 3 number of possibilities together for the total amount for List order of boy, girl, boy.

Repeat similarly for the second list order.

Just worked it out:

List 1 = (12 × 10 × 11)

List 2 = (10 × 12 × 9)

List 1 + List 2 = 2400 list possibilities.

Thanks for your help

Original post by Adidas02

Just worked it out:

List 1 = (12 × 10 × 11)

List 2 = (10 × 12 × 9)

List 1 + List 2 = 2400 list possibilities.

Thanks for your help

List 1 = (12 × 10 × 11)

List 2 = (10 × 12 × 9)

List 1 + List 2 = 2400 list possibilities.

Thanks for your help

If you need more practice, there are exam questions here.

Original post by Notnek

I haven't checked your working but please don't post full solutions - post hints instead.

Please read the posting guidelines.

Please read the posting guidelines.

I am so so sorry, I'm new to this and didn't know.

My apologies

Original post by Notnek

I haven't checked your working but please don't post full solutions - post hints instead.

Please read the posting guidelines.

Please read the posting guidelines.

They didn't seem to read the question properly anyway.

Original post by Notnek

If you need more practice, there are exam questions here.

Thank you

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