User interface language: English | Español

Date November 2014 Marks available 4 Reference code 14N.1.hl.TZ0.10
Level HL only Paper 1 Time zone TZ0
Command term Find Question number 10 Adapted from N/A

Question

A set of positive integers {1,2,3,4,5,6,7,8,91,2,3,4,5,6,7,8,9} is used to form a pack of nine cards.

Each card displays one positive integer without repetition from this set. Grace wishes to select four cards at random from this pack of nine cards.

Find the number of selections Grace could make if the largest integer drawn among the four cards is either a 55, a 66 or a 77.

[3]
a.

Find the number of selections Grace could make if at least two of the four integers drawn are even.

[4]
b.

Markscheme

use of the addition principle with 33 terms     (M1)

to obtain 4C3+5C3+6C3 (=4+10+20)4C3+5C3+6C3 (=4+10+20)     A1

number of possible selections is 3434     A1

[3 marks]

a.

EITHER

recognition of three cases: (22 odd and 22 even or 11 odd and 33 even or 00 odd and 44 even)     (M1)

(5C2×4C2)+(5C1×4C3)+(5C0×4C4)(=60+20+1)(5C2×4C2)+(5C1×4C3)+(5C0×4C4)(=60+20+1)     (M1)A1

OR

recognition to subtract the sum of 44 odd and 33 odd and 11 even from the total     (M1)

9C45C4(5C3×4C1)(=126540)9C45C4(5C3×4C1)(=126540)     (M1)A1

THEN

number of possible selections is 8181     A1

[4 marks]

Total [7 marks]

b.

Examiners report

As the last question on section A, candidates had to think about the strategy for finding the answers to these two parts. Candidates often had a mark-worthy approach, in terms of considering separate cases, but couldn’t implement it correctly.

a.

As the last question on section A, candidates had to think about the strategy for finding the answers to these two parts. Candidates often had a mark-worthy approach, in terms of considering separate cases, but couldn’t implement it correctly.

b.

Syllabus sections

Topic 1 - Core: Algebra » 1.3 » Counting principles, including permutations and combinations.

View options