PnC Qns and Solns (SAJC)

Chapter 1: Permutation and Combination 1. HCI JC2 Prelim 8865/2019/Q6 (i) Janet is going on a holiday trip and is trying to recall the 4-digit numerical combination lock for her luggage. However, she can only remember the numerical combination consists of digits from the set { 0, 1, 4, 6, 8, 9} . (a) How many different numerical combinations are there if the odd and even digits must alternate and repetitions of digits are not allowed? [1] (b) If the 4-digit numerical combination must be a number that is more than 6400 and that there are no repetitions of digits, find the maximum number of numerical combinations Janet has to try in order to open her lock? [2] (ii) Janet has 6 T-shirts, 3 sweaters and 3 pairs of jeans, all of which are of different designs. Find the number of ways she can select her clothes for her trip if she needs 4 T -shirts, a sweater and at least a pair of jeans. [3] Answer: (ia) 48, (ib) 156, (ii) 315 6. HCI JC2 Prelim 8865/2019/Q6 (Solutions) (ia) Set of digits to choose from = {0, 1, 4, 6, 8, 9} Note: Zero is even. Case 1: odd even odd even = 2 4 1 3 24   = Case 2: even odd even odd = 4 2 3 1 24   = No. of combinations = ( )2 4 3 2 1 48    = (ib) Case 1: 6 _ _ _ 4 or 8 or 9 No. of combinations = 1 3 4 3 36   = Case 2: _ _ _ _ 8 or 9 No. of combinations = 2 5 4 3 120   = Total no. of combinations 36 120 156+= (ii) 6T, 3J, 3W No. of ways to select T-shirts = 6 154  =

No. of ways to select jeans = 333 71 2 3      ++=           No. of ways to select sweater = 3 Total no. of ways = 15 7 3 315  = 2. NJC JC2 Prelim 8865/2019/Q8 Find the number of ways in which the letters of the word QUESTION can be arranged, if (i) Q and U must be next to each other, [2] (ii) E and S must be separated, [2] (iii) consonants (Q, S, T, N) and vowels (U, E, I, O) must alternate. [2] A code consists of 5 letters chosen from {Q, U, E, S, T, I, O, N}. Repetitions are not allowed. (iv) Find the probability that a randomly chosen code has only 1 vowel. [2] Answer: (i) 10080, (ii) 30240, (iii) 1152, (iv) 1 14 2. NJC JC2 Prelim 8865/2019/Q8 (Solutions) (i) No. of ways to arrange Q and U 2!= No. of ways to arrange all letters with Q and U grouped together 7! 5040== No. of ways that Q and U must be next to each other 7!2! 10080== (ii) Method 1: Slotting No. of ways to arrange the other 6 letters 6! 720== No. of ways to choose slots and arrange E and S 7 2! 422 =  = No. of ways E and S must be separated 76! 2! 302402 =   = (ii) Method 2: Complementary No. of ways to arrange all the 8 letters 8! 40320== No. of ways E and S must be separated 8! (i) 30240= − = (iii) To alternate, the pattern must be CVCVCVCV or VCVCVCVC, where C = consonant, V = vowel. No. of ways to arrange the consonants 4! 24== No. of ways to arrange the vowels 4! 24== No. of ways to alternate 2 4!4! 1152=  = (iv) Method 1: Using Probability

Probability (consider 5 cases to arrange VCCCC) 4 4 3 2 1 1 58 7 6 5 4 14=      = Method 2: Using P&C No. of ways to have 1 vowel 4 5! 4801 =  = Total number of 5-letter codes 8 5! 67205 =  = Probability 480 1 6720 14== 3. ACJC JC2 Prelim 8865/2019/Q7 A certain restaurant employs 4 cashiers, 5 ushers and 6 food servers as full -time service staff. A shift consists of a team of 2 cashiers, 4 ushers and 4 food servers on duty. (i) How many different teams can be formed to work in a shift? [1] (ii) Out of the 6 food servers, there is a pair of sisters working together as food servers in the restaurant. How many different selections of food servers are there in order to include exactly one of the two sisters in a shift? [1] (iii) The restaurant also employs a part-timer who can work either as a cashier or as an usher. How many different ways can a shift be formed that must include this particular part-timer?[3] (iv) All 15 full-time service staff gather in a straight line for a briefi ng. Find the number of ways the line can be formed such that all 5 ushers stand alternately with all 6 food servers. [2] Answer: (i) 450, (ii) 8, (iii) 1200, (iv) 10368000 3. ACJC JC2 Prelim 8865/2019/Q7 (Solutions) (i) 456 2 4 4No. of teams 6 5 15 450C C C=   =   = (ii) 24 13 No. of selections of food servers to include exactly 1 sister 2 4 8CC=  =  = (iii) Case 1: Part-timer as cashier 4 5 6 1 4 4No. of cashier-usher teams 4 5 15 300C C C=   =   = Case 2: Part-timer as usher 4 5 6 2 3 4No. of cashier-usher teams 6 10 15 900C C C=   =   = No. of teams 300 900 1200 = + = (iv) No. of ways 5! 6! 5! 120 720 120 10,3 68,000 =   =   = 4. TMJC JC2 Prelim 8865/2019/Q6

A company has 10 software engineers, 2 civil engineers and 5 electrical engineers. The company would like to send a delegation of 8 engineers for a conference. (i) How many different delegations can be formed if at least 2 of each type of engineers must be included in the delegation for the conference? [3] It is given that the delegation of 8 engineers for the conference must include 5 software engineers, 1 civil engineer and 2 electrical engineers. (ii) How many different delegations can be formed by the company? [1] (iii) One of the software engineers in the company is the brother of one of the electrical engineers in the company. How many different delegations can be formed which include exactly one of the two brothers? [3] Answer: (i) 3525, (ii) 5040, (iii) 2520 4. TMJC JC2 Prelim 8865/2019/Q6 (Solutions) (i) Number of delegations formed = (3 software, 2 civil, 3 electrical) + (4 software, 2 civil, 2 electrical) + (2 software, 2 civil, 4 electrical) = ( ) ( ) ( ) 10 2 5 10 2 5 10 2 5 3 2 3 4 2 2 2 2 4 3525C C C C C C C C C  +   +   = (ii) Number of delegations = 10 2 5 512 5040C C C  = (iii) Case 1: only software engineer brother is in Number of delegations = 9 2 4 4 1 21 1512C C C   = Case 2: only electrical engineer brother is in Number of delegations = 9 2 4 5 1 1 1 1008C C C   = Total number of delegations formed = 1512 + 1008 = 2520 5. ASRJC JC2 Prelim 8865/2019/Q6 A game is played with 12 cards, consisting of 4 colours , Red, Green, Blue, and Yellow. Each set of colour consists of 3 cards labelled X, Y, and Z. The 12 cards are arranged in a row. (a) (i) How many different ways can the 12 cards be arranged so that the 3 cards in each coloured set are next to each other? [2] (ii) How many different ways can the cards be arranged so that not all Red cards are together? [2]

(iii) How many different ways can the cards be arranged so that all four cards labelled X are next to each other and all three Green cards are next to each other? [3] (b) The cards are shuffled and arranged in a row. Given that the first card is labelled Z, find the exact probability that no two cards labelled Z are next to each other. [4] Answer: (ai) 31104, (aii) 457 228 800, (aiii) 120960, (b) 56 165 5. ASRJC JC2 Prelim 8865/2019/Q6 (Solutions) (a) (i) No. of Ways ( ) 4 4!