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Step-by-step solution for: Solved What is the title of This Picture? 12 Find each | Chegg.com
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Step-by-step solution for: Solved What is the title of This Picture? 12 Find each | Chegg.com
Let’s solve each problem one by one. We’ll use combinations and permutations where needed. Remember:
- Combinations (C): Order doesn’t matter → use when selecting groups, teams, toppings, etc.
Formula: \( nCr = \frac{n!}{r!(n-r)!} \)
- Permutations (P): Order matters → use when arranging in order, like positions on council, race finishes, etc.
Formula: \( nPr = \frac{n!}{(n-r)!} \)
---
G. 5C3
This means “choose 3 out of 5”, order doesn’t matter.
\( 5C3 = \frac{5!}{3!(5-3)!} = \frac{5×4×3!}{3! × 2!} = \frac{5×4}{2×1} = 10 \)
✔ Answer: 10
---
I. 9C2
Choose 2 out of 9.
\( 9C2 = \frac{9×8}{2×1} = 36 \)
✔ Answer: 36
---
H. 7C4
Choose 4 out of 7.
Note: \( 7C4 = 7C3 \) because choosing 4 to include is same as choosing 3 to leave out.
\( 7C4 = \frac{7×6×5×4}{4×3×2×1} = \frac{840}{24} = 35 \)
Or faster: \( 7C3 = \frac{7×6×5}{3×2×1} = 35 \)
✔ Answer: 35
---
O. 20C19
Choose 19 out of 20 → same as leaving out 1 person → so 20 ways.
\( 20C19 = 20 \)
✔ Answer: 20
---
Now the word problems:
---
S. How many committees of 4 members can be formed from 9 people?
Order doesn’t matter → combination.
\( 9C4 = \frac{9×8×7×6}{4×3×2×1} = \frac{3024}{24} = 126 \)
✔ Answer: 126
---
E. In how many ways can two flavors of ice cream be selected from 31 flavors?
Again, order doesn’t matter — vanilla + chocolate is same as chocolate + vanilla.
\( 31C2 = \frac{31×30}{2} = 465 \)
✔ Answer: 465
---
R. Lincoln High School Orchestra has 8 pieces. Needs to play 3 for competition. Same pieces in different order = different program.
→ Order matters → permutation!
\( 8P3 = 8×7×6 = 336 \)
✔ Answer: 336
---
N. Same orchestra, but now same pieces in different order = SAME selection.
→ Order doesn’t matter → combination.
\( 8C3 = \frac{8×7×6}{3×2×1} = 56 \)
✔ Answer: 56
---
K. Seven students running for two positions: first and second place.
Order matters → who gets first vs second is different.
So it’s permutation: \( 7P2 = 7×6 = 42 \)
✔ Answer: 42
---
A. Seven students running for two positions. How many different PAIRS could be elected?
“Pairs” implies order doesn’t matter — just which two are chosen, not who is first or second.
So combination: \( 7C2 = \frac{7×6}{2} = 21 \)
✔ Answer: 21
---
F. Student must answer any four of six essay questions. How many different selections?
Order doesn’t matter — just which 4 they pick.
\( 6C4 = 6C2 = \frac{6×5}{2} = 15 \)
✔ Answer: 15
---
M. Pico Blox bookmarked 12 websites. Visits five. How many different combinations?
Order doesn’t matter — just which 5 he visits.
\( 12C5 = \frac{12×11×10×9×8}{5×4×3×2×1} = \frac{95040}{120} = 792 \)
✔ Answer: 792
---
L. PizzaZZZ offers ten toppings. Choose three. How many combinations?
\( 10C3 = \frac{10×9×8}{3×2×1} = 120 \)
✔ Answer: 120
---
W. Two members chosen from 400 audience for TV quiz show. How many different selections?
Assuming order doesn’t matter — just which two are picked.
\( 400C2 = \frac{400×399}{2} = 79800 \)
✔ Answer: 79800
---
T. How many different 6-player starting squads from 11 players?
Order doesn’t matter — squad is a group.
\( 11C6 = 11C5 = \frac{11×10×9×8×7}{5×4×3×2×1} = \frac{55440}{120} = 462 \)
✔ Answer: 462
---
C. How many different 5-card hands from standard deck of 52 cards?
Classic combo problem.
\( 52C5 = \frac{52×51×50×49×48}{5×4×3×2×1} = \frac{311875200}{120} = 2,598,960 \)
✔ Answer: 2,598,960
---
Final Answers Summary:
G. 5C3 → 10
I. 9C2 → 36
H. 7C4 → 35
O. 20C19 → 20
S → 126
E → 465
R → 336
N → 56
K → 42
A → 21
F → 15
M → 792
L → 120
W → 79800
T → 462
C → 2,598,960
──────────────────────────────────────
Final Answer:
G: 10, I: 36, H: 35, O: 20, S: 126, E: 465, R: 336, N: 56, K: 42, A: 21, F: 15, M: 792, L: 120, W: 79800, T: 462, C: 2598960
- Combinations (C): Order doesn’t matter → use when selecting groups, teams, toppings, etc.
Formula: \( nCr = \frac{n!}{r!(n-r)!} \)
- Permutations (P): Order matters → use when arranging in order, like positions on council, race finishes, etc.
Formula: \( nPr = \frac{n!}{(n-r)!} \)
---
G. 5C3
This means “choose 3 out of 5”, order doesn’t matter.
\( 5C3 = \frac{5!}{3!(5-3)!} = \frac{5×4×3!}{3! × 2!} = \frac{5×4}{2×1} = 10 \)
✔ Answer: 10
---
I. 9C2
Choose 2 out of 9.
\( 9C2 = \frac{9×8}{2×1} = 36 \)
✔ Answer: 36
---
H. 7C4
Choose 4 out of 7.
Note: \( 7C4 = 7C3 \) because choosing 4 to include is same as choosing 3 to leave out.
\( 7C4 = \frac{7×6×5×4}{4×3×2×1} = \frac{840}{24} = 35 \)
Or faster: \( 7C3 = \frac{7×6×5}{3×2×1} = 35 \)
✔ Answer: 35
---
O. 20C19
Choose 19 out of 20 → same as leaving out 1 person → so 20 ways.
\( 20C19 = 20 \)
✔ Answer: 20
---
Now the word problems:
---
S. How many committees of 4 members can be formed from 9 people?
Order doesn’t matter → combination.
\( 9C4 = \frac{9×8×7×6}{4×3×2×1} = \frac{3024}{24} = 126 \)
✔ Answer: 126
---
E. In how many ways can two flavors of ice cream be selected from 31 flavors?
Again, order doesn’t matter — vanilla + chocolate is same as chocolate + vanilla.
\( 31C2 = \frac{31×30}{2} = 465 \)
✔ Answer: 465
---
R. Lincoln High School Orchestra has 8 pieces. Needs to play 3 for competition. Same pieces in different order = different program.
→ Order matters → permutation!
\( 8P3 = 8×7×6 = 336 \)
✔ Answer: 336
---
N. Same orchestra, but now same pieces in different order = SAME selection.
→ Order doesn’t matter → combination.
\( 8C3 = \frac{8×7×6}{3×2×1} = 56 \)
✔ Answer: 56
---
K. Seven students running for two positions: first and second place.
Order matters → who gets first vs second is different.
So it’s permutation: \( 7P2 = 7×6 = 42 \)
✔ Answer: 42
---
A. Seven students running for two positions. How many different PAIRS could be elected?
“Pairs” implies order doesn’t matter — just which two are chosen, not who is first or second.
So combination: \( 7C2 = \frac{7×6}{2} = 21 \)
✔ Answer: 21
---
F. Student must answer any four of six essay questions. How many different selections?
Order doesn’t matter — just which 4 they pick.
\( 6C4 = 6C2 = \frac{6×5}{2} = 15 \)
✔ Answer: 15
---
M. Pico Blox bookmarked 12 websites. Visits five. How many different combinations?
Order doesn’t matter — just which 5 he visits.
\( 12C5 = \frac{12×11×10×9×8}{5×4×3×2×1} = \frac{95040}{120} = 792 \)
✔ Answer: 792
---
L. PizzaZZZ offers ten toppings. Choose three. How many combinations?
\( 10C3 = \frac{10×9×8}{3×2×1} = 120 \)
✔ Answer: 120
---
W. Two members chosen from 400 audience for TV quiz show. How many different selections?
Assuming order doesn’t matter — just which two are picked.
\( 400C2 = \frac{400×399}{2} = 79800 \)
✔ Answer: 79800
---
T. How many different 6-player starting squads from 11 players?
Order doesn’t matter — squad is a group.
\( 11C6 = 11C5 = \frac{11×10×9×8×7}{5×4×3×2×1} = \frac{55440}{120} = 462 \)
✔ Answer: 462
---
C. How many different 5-card hands from standard deck of 52 cards?
Classic combo problem.
\( 52C5 = \frac{52×51×50×49×48}{5×4×3×2×1} = \frac{311875200}{120} = 2,598,960 \)
✔ Answer: 2,598,960
---
Final Answers Summary:
G. 5C3 → 10
I. 9C2 → 36
H. 7C4 → 35
O. 20C19 → 20
S → 126
E → 465
R → 336
N → 56
K → 42
A → 21
F → 15
M → 792
L → 120
W → 79800
T → 462
C → 2,598,960
──────────────────────────────────────
Final Answer:
G: 10, I: 36, H: 35, O: 20, S: 126, E: 465, R: 336, N: 56, K: 42, A: 21, F: 15, M: 792, L: 120, W: 79800, T: 462, C: 2598960
Parent Tip: Review the logic above to help your child master the concept of what are the titles of these pictures math worksheet answers.
