Choosing an Outfit
3 pairs of pants (B, R, O)
2 tops (W, R)
2 shoes (B, W)
- Tree Diagram
# of outfits = 3 × 2 × 2 = 12
2.How many 3-digit numbers can be formed if:
- there are no restrictions? ANS: 9 × 10 × 10 = 900 (a 3 digit number can’t start with zero)
- there are no repetitions? ANS: 9 × 9 × 8 = 648 (gain the zero, lose the 1st digit number)
- How many 4-letter arrangements can be formed if: (26 letters in the alphabet?)
- there are no repetitions? ANS: 26 × 25 × 24 × 23 = 358 800
- repetitions allowed? ANS: 264 = 456 976 a.k.a. 26 x 26 x 26 x 26 =
- How many numbers can be formed from the digits 1, 4, 6, and 8 (one of each) if the number:
- Must be even ANS: 3 × 2 × 1 × 3 (do first) = 18
- Must be greater than 5 000. ANS: 2 × 3 × 2 × 1 = 12
- Must be greater than 400. ANS: 3 × 3 × 2 + 4 × 3 × 2 × 1 = 18 + 24 = 42
- How many 5-letter arrangements can be made from the letters A, B, C, D, E, F, G?
ANS: 7P5 or 7 × 6 × 5 × 4 × 3 = 2 520
- How many ways can 6 books be arranged on a shelf?
ANS: 6P6 or 6 × 5 × 4 × 3 × 2 × 1 = 720
- 5-letter arrangements are formed from the letters in NUMBERS. How many if:
- no vowels? ANS: 5 × 4 × 3 × 2 × 1 = 120
- 1st and last letters are consonants? ANS: 5 × 5 × 4 × 3 × 4 = 1 200
___ x ___ x ___ x ___ x ____ AEIOU are vowels in the alphabet
Con Any Any Any Con
- 6-letter arrangements are formed from ABCDEF. How many if:
- no restrictions?
ANS: 6 × 5 × 4 × 3 × 2 × 1 = 6! = 720
- A & B together? (think of AB as one letter)
ANS: 5 × 4 × 3 × 2 × 1 × 2 × 1 = 120
Number of ways to arrange AB ↵
The 2 × 1 is the placing of the A and B
The 5 × 4 × 3 × 2 × 1 are the placing of the AB as one unit and the CDEF as 4 units.
- A, B, & C together? ANS: 4 × 3 × 2 × 1 × 3 × 2 × 1 = 144
The 4 × 3 × 2 × 1 is the arranging of DEF (in any order) and the ABC as a whole.
The 3 × 2 × 1 is the arranging of the ABC.
- A & B separated? (arrange others first) ANS: 4 × 3 × 2 × 1 × 5 × 4 = 480 OR #a – #b
The 4 × 3 × 2 × 1 is the CDEF
The 5 is the A or B (one or the other) and the CDEF
You basically have __ C __ D __ E __ F ___
Now the CDEF can go in any order hence the 4!
The A or B could go in any of the 5 spaces hence the multiplying 5
Once the A or B is placed, the other one has 4 difference places to go, hence the multiplying 4.
- A, B, & C separated? ANS: 3 × 2 × 1 × 4 × 3 × 2 = 144 (can’t do as above)
Same hint as above. You basically have __ D __ E __ F ___
Now the DEF can go in any order hence the 4!
The A or B or C could go in any of the 4 spaces hence the multiplying 4
Once the A B or C is placed, the other two have 3 difference places to go, hence the multiplying 3.
The multiplying 2 is the last A, B or C to go in either of the empty two spots.
- the word BAD appears? ANS: 4 × 3 × 2 × 1 = 24
The 4 is the word BAD or C, E, or F (4 items as BAD is one unit)
The 3 is the 4 options above less the one already used.
The 2 is the 3 options above less the one already used.
The 1 is the only option left.
BAD cannot be switched as it must remain as BAD
- How many arrangements of the letters in the word:
- BANANA
ANS: 3 A’s and 2 N’s
- STATISTICS
ANS: 3 S’s, 3 T’s and 2 I’s
Computation
A committee of 3 is selected from 5 girls and 7 boys. How many if:
- no restrictions? ANS: 12C3 = 220
- all boys? ANS: 7C3 = 35
- 1 girl and 2 boys? ANS: 5C1 ∙ 7C2 = 105
- At least 1 girl? ANS: 1 girl or 2 girls or 3 girls = 5C1 ∙ 7C2 + 5C2 ∙ 7C1 + 5C3
= 105 + 70 + 10
= 185
OR #a − #b = 220 − 35
= 185
- 10 points are on a circle (no 3 are collinear). How many:
- Different lines are determined?
ANS: 10C2 = 45
A line is a connection of 2 points. So, take the 10 points and connect 2 of them.
- quadrilaterals?
ANS: 10C4 = 210
A quadrilateral has 4 sides. So, take 10 points and connect 4 of them.