MEI S1 Ex 5B Q12 Help please!

Hi everyone, you will probably only be able to help me with this if you have the MEI S1 textbook. The questions is, as in the title, Exercise 5B - Question 12. I can do the first two parts with no trouble but beyond that not much of it makes sense.

For part iii I looked at the answers in the back and thankfully they reasoned their answer. However for the next bit I then tried to apply their logic to a similar question, producing a wrong answer. I cannot see why 5C3 x 22 x 21 is a correct answer here, seeing as it applies the same logic as the last part.

Hi everyone, you will probably only be able to help me with this if you have the MEI S1 textbook. The questions is, as in the title, Exercise 5B - Question 12. I can do the first two parts with no trouble but beyond that not much of it makes sense.

For part iii I looked at the answers in the back and thankfully they reasoned their answer. However for the next bit I then tried to apply their logic to a similar question, producing a wrong answer. I cannot see why 5C3 x 22 x 21 is a correct answer here, seeing as it applies the same logic as the last part.

Thank you for your help, I see how you get those answers now, and in part (iv) I did mean £10, sorry about that. Thanks again.

Bit confused as to which parts you're looking at. So here's solutions to first 4.

i) Largest number of tickets, since they are all different, is from 26 choose 5, so: $^{26}C_5$

ii) Clearly $\dfrac{1}{^{26}C_5}$


iii) I'm sure you've got your own examples.

Then to win £100 you need 4 winning letters, and 1 that doesn't win. So chose 4 from 5 winning ones, and then 1 from the remaining 21 losing letters.

i.e. $^{5}C_4\times ^{21}C_1= 5\times 21 = 105$

iv) The book is wrong here, IMHO.
I presume you mean tickets that win £10, rather than £100.

So you need 3 winning letters, and 2 losing letters.
I.e. from 5 choose 3, and from 21 choose 2.

$^5C_3\times ^{21}C_2= ^5C_3 \times 21\times 20 /2=2100$

Damn, your logic here is brilliantly simple! Why didn't I think of that Thanks for the help, I was also stuck on this question.

Bit confused as to which parts you're looking at. So here's solutions to first 4.

i) Largest number of tickets, since they are all different, is from 26 choose 5, so: $^{26}C_5$

ii) Clearly $\dfrac{1}{^{26}C_5}$


iii) I'm sure you've got your own examples.

Then to win £100 you need 4 winning letters, and 1 that doesn't win. So chose 4 from 5 winning ones, and then 1 from the remaining 21 losing letters.

i.e. $^{5}C_4\times ^{21}C_1= 5\times 21 = 105$

iv) The book is wrong here, IMHO.
I presume you mean tickets that win £10, rather than £100.

So you need 3 winning letters, and 2 losing letters.
I.e. from 5 choose 3, and from 21 choose 2.

$^5C_3\times ^{21}C_2= ^5C_3 \times 21\times 20 /2=2100$

I also searched for this and found it

I don't understand the logic for part iv) though - I thought the number of ways of winning £10 would be 5C3 x 23 x 22 (since once you've got the 3 letters the same, there would be 23 possible letters for a fourth and 22 for the fifth) Where is the 21 and 20/2 coming from?