Math  /  Discrete

QuestionMDM4U1 - Data Man. Name: Date: Ch 5, Sec. 5.2 to 5.3 Assign 15 K 18 T 15 C (overall) 16 A Full marks are only given to solutions that are Complete, Clear and Correct Q1) There are 8 boys and 12 girls in a class. a) [2K] Determine the number of different groups that contain exactly four girls and five boys. b) [3K] How many groups of 14 have an equal number of boys and girls? c) [2T] How many groups of five have more boys than girls?
Q2) If you are dealing from a standard deck of 52 playing cards: a) [2 A][2 \mathrm{~A}] How many different 5 -card hands could have at least two spades? b) [2A] How many different 5-card hands could have at most two court/picture cards? [Note: Court/picture cards are the jacks, queens and kings.]

Studdy Solution

STEP 1

1. We are dealing with combinations since the order of selection does not matter.
2. A standard deck of cards has 52 cards, with 13 cards in each suit and 12 court cards in total (3 per suit).

_HIGH_LEVEL_APPROACH_ for Q1:
1. Calculate the number of groups with exactly four girls and five boys.
2. Calculate the number of groups of 14 with an equal number of boys and girls.
3. Calculate the number of groups of five with more boys than girls.

_HIGH_LEVEL_APPROACH_ for Q2:
1. Calculate the number of 5-card hands with at least two spades.
2. Calculate the number of 5-card hands with at most two court cards.

**Q1:**

STEP 2

STEP 3

Calculate the number of groups with exactly four girls and five boys.
- Use the combination formula: (nr) \binom{n}{r} , where n n is the total number of items, and r r is the number of items to choose. - Calculate the number of ways to choose 4 girls from 12: (124) \binom{12}{4} . - Calculate the number of ways to choose 5 boys from 8: (85) \binom{8}{5} . - Multiply the results to get the total number of groups.
(124)=12×11×10×94×3×2×1=495 \binom{12}{4} = \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2 \times 1} = 495 (85)=8×7×63×2×1=56 \binom{8}{5} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 Total groups=495×56=27720 \text{Total groups} = 495 \times 56 = 27720

STEP 4

Calculate the number of groups of 14 with an equal number of boys and girls.
- Choose 7 boys from 8: (87) \binom{8}{7} . - Choose 7 girls from 12: (127) \binom{12}{7} . - Multiply the results to get the total number of groups.
(87)=8 \binom{8}{7} = 8 (127)=12×11×10×9×85×4×3×2×1=792 \binom{12}{7} = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792 Total groups=8×792=6336 \text{Total groups} = 8 \times 792 = 6336

STEP 5

Calculate the number of groups of five with more boys than girls.
- Possible combinations: 3 boys, 2 girls or 4 boys, 1 girl or 5 boys. - Calculate each scenario and sum them.
For 3 boys, 2 girls: (83)×(122)=56×66=3696 \binom{8}{3} \times \binom{12}{2} = 56 \times 66 = 3696
For 4 boys, 1 girl: (84)×(121)=70×12=840 \binom{8}{4} \times \binom{12}{1} = 70 \times 12 = 840
For 5 boys: (85)=56 \binom{8}{5} = 56
Total groups: 3696+840+56=4592 3696 + 840 + 56 = 4592
**Q2:**
STEP_1: Calculate the number of 5-card hands with at least two spades.
- Calculate total 5-card hands: (525) \binom{52}{5} . - Calculate hands with 0 or 1 spade and subtract from total.
Total hands: (525)=2598960 \binom{52}{5} = 2598960
Hands with 0 spades: (395)=575757 \binom{39}{5} = 575757
Hands with 1 spade: (131)×(394)=13×82251=1069263 \binom{13}{1} \times \binom{39}{4} = 13 \times 82251 = 1069263
Hands with at least 2 spades: 2598960(575757+1069263)=953940 2598960 - (575757 + 1069263) = 953940
STEP_2: Calculate the number of 5-card hands with at most two court cards.
- Calculate hands with 0, 1, or 2 court cards and sum them.
Hands with 0 court cards: (405)=658008 \binom{40}{5} = 658008
Hands with 1 court card: (121)×(404)=12×91,390=1,096,680 \binom{12}{1} \times \binom{40}{4} = 12 \times 91,390 = 1,096,680
Hands with 2 court cards: (122)×(403)=66×9880=651,840 \binom{12}{2} \times \binom{40}{3} = 66 \times 9880 = 651,840
Total hands: 658008+1,096,680+651,840=2,406,528 658008 + 1,096,680 + 651,840 = 2,406,528
The solutions are: - Q1a: 27720 27720 - Q1b: 6336 6336 - Q1c: 4592 4592 - Q2a: 953940 953940 - Q2b: 2406528 2406528

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