22M.2.AHL.TZ2.9

pestleMathematicsAAHLPaper 222M· ahl-1-10-perms-and-combs-binomial-with-negative-and-fractional-indicessource ↗

Consider the set of six-digit positive integers that can be formed from the digits $0, 1, 2, 3, 4, 5, 6, 7, 8$ and $9$ .

Find the total number of six-digit positive integers that can be formed such that

the digits are distinct.

[2]

the digits are distinct and are in increasing order.

[2]

Markscheme / solution

$9 \times 9 \times 8 \times 7 \times 6 \times 5 (= 9 \times {}_{9}P_{5})$ (M1)

$= 136080 (= 9 \times \frac{9!}{4!})$ A1

Note: Award M1A0 for $10 \times 9 \times 8 \times 7 \times 6 \times 5 (= {}_{10}P_{6} = 151200 = \frac{10!}{4!})$ .

Note: Award M1A0 for ${}_{9}P_{6} = 60480$

[2 marks]

METHOD 1

EITHER

every unordered subset of $6$ digits from the set of $9$ non-zero digits can be arranged in exactly one way into a $6$ -digit number with the digits in increasing order. A1

${}^{9}C_{6} (\times 1)$ A1

THEN

$= 84$ A1

METHOD 2

EITHER

removes $3$ digits from the set of $9$ non-zero digits and these $6$ remaining digits can be arranged in exactly one way into a $6$ -digit number with the digits in increasing order. A1

${}^{9}C_{3} (\times 1)$ A1

THEN

$= 84$ A1

[2 marks]

Examiners’ report

Part (a) A number of candidates got the correct answer here with the valid approach and recognising that zero could not occupy the first position. Some lost a mark by including zero. Many candidates used an incorrect method with combinations or simplified permutations.

Part (b) Only very few candidates got the correct answer. Many left it blank or provided unreasonably enormous numbers as their answers.

Some candidates had the answer to part (b) showing in part (a).

A small number of candidates tried to list all possibilities but mostly unsuccessfully.

[N/A]