Systematic lists and combinations

(a) CT, CJ, HT, HJ, ET, EJ

1. Work through the sandwiches in order, pairing each with every drink: CT, CJ, then HT, HJ, then ET, EJ.

• B1 At least three correct combinations with no repeats.
• B1 All six combinations, with no extras and no repeats.

(b) $6$

1. Count the list: $3$ sandwiches each with $2$ drinks makes $6$.

• B1 The correct answer, $6$.

(a) $6$

1. Pair each coin only with the coins after it in the list: 10p with 20p, 50p, £1; 20p with 50p, £1; 50p with £1.
2. $3 + 2 + 1 = 6$ pairs.

• M1 A systematic list of pairs with at least four correct and no repeats.
• A1 The correct answer, $6$.

(b) $3$

1. The totals are 30p, 60p, 110p, 70p, 120p and 150p.
2. Those over 75p are 110p, 120p and 150p: $3$ pairs, each containing the £1 coin.

• M1 Finding the totals of the pairs.
• A1 The correct answer, $3$.

(a) $4$

1. An even number must end in 2 or 4. List the possibilities by last digit.
2. An even number must end in 2 or 4. List the possibilities by last digit.
3. Ending in 2 gives 42 and 72. Ending in 4 gives 24 and 74.
4. There are 4 different numbers.

• P1 An even number must end in 2 or 4. List the possibilities by last digit.
• A1 Correct answer: $4$

(a) $16$

1. Count all unrestricted choices.
   $$3\times 3\times 2$$
2. Remove the two drinks paired with the unavailable sandwich.
   $$18-2$$
3. Therefore $16$.

• P1 Count all unrestricted choices.
• P1 Remove the two drinks paired with the unavailable sandwich.
• A1 Correct answer: $16$

(b) Each filling can be paired with each bread and each drink, so combinations require multiplication before removing restrictions.

1. Each filling can be paired with each bread and each drink, so combinations require multiplication before removing restrictions.

• C1 Correct conclusion with supporting reasoning: Each filling can be paired with each bread and each drink, so combinations require multiplication before removing restrictions.

(a) $16$

1. Each even units digit leaves one fewer tens choices.
   $$4\times 4$$
2. Therefore $16$.

• P1 Each even units digit leaves one fewer tens choices.
• A1 Correct answer: $16$

(b) $4$

1. The units digit must be 1; choose any other tens digit.
   $$5-1$$
2. Therefore $4$.

• M1 The units digit must be 1; choose any other tens digit.
• A1 Correct answer: $4$

(a) $15$

1. Count codes beginning with B or C.
   $$2\times 6$$
2. Add the choices for A using only the even digits.
   $$12+3$$
3. Therefore $15$.

• P1 Count codes beginning with B or C.
• P1 Add the choices for A using only the even digits.
• A1 Correct answer: $15$

(b) List all A codes in digit order, then all B codes, then all C codes; apply the even-digit restriction only to A.

1. List all A codes in digit order, then all B codes, then all C codes; apply the even-digit restriction only to A.

• C1 Correct conclusion with supporting reasoning: List all A codes in digit order, then all B codes, then all C codes; apply the even-digit restriction only to A.

(a) $9$

1. Each even units digit leaves one fewer tens choices.
   $$3\times 3$$
2. Therefore $9$.

• P1 Each even units digit leaves one fewer tens choices.
• A1 Correct answer: $9$

(b) $3$

1. The units digit must be 1; choose any other tens digit.
   $$4-1$$
2. Therefore $3$.

• M1 The units digit must be 1; choose any other tens digit.
• A1 Correct answer: $3$

(a) $12$

1. List with the numbers in decreasing order, starting from the largest possible first number.
2. $$10{+}1{+}1,\ 9{+}2{+}1,\ 8{+}3{+}1,\ 8{+}2{+}2,\ 7{+}4{+}1,\ 7{+}3{+}2,$$
3. $$6{+}5{+}1,\ 6{+}4{+}2,\ 6{+}3{+}3,\ 5{+}5{+}2,\ 5{+}4{+}3,\ 4{+}4{+}4$$
4. That is 12 sets.

• P1 A systematic method: writing each set in decreasing order and working down from the largest first number.
• P1 At least eight correct sets with no repeats.
• A1 The correct answer, $12$.

(b) $3$

1. A triangle is possible only when the longest side is less than the sum of the other two.
2. With a total of $12$, the longest side $L$ needs $L < 12 - L$, so $L < 6$.
3. The sets with largest number below $6$: $5 + 5 + 2$, $5 + 4 + 3$, $4 + 4 + 4$. That is $3$.

• P1 Using the triangle condition, largest side less than the sum of the other two (so less than 6).
• A1 The correct answer, $3$.