# General Statistics probability calculations [Step by Step]

Choosing an Outfit

3 pairs of pants (B, R, O)
2 tops (W, R)
2 shoes (B, W)

1.     Tree Diagram

# of outfits = 3 × 2 × 2 = 12

2.How many 3-digit numbers can be formed if:

1.     there are no restrictions?        ANS: 9 × 10 × 10 = 900 (a 3 digit number can’t start                         with zero)
2.     there are no repetitions?        ANS: 9 × 9 × 8 = 648 (gain the zero, lose the 1st digit                         number)
3. How many 4-letter arrangements can be formed if:    (26 letters in the alphabet?)
4.     there are no repetitions?        ANS: 26 × 25 × 24 × 23 = 358 800
5.     repetitions allowed?        ANS: 264 = 456 976 a.k.a. 26 x 26 x 26 x 26 =
6. How many numbers can be formed from the digits 1, 4, 6, and 8 (one of each) if the         number:
7. Must be even     ANS: 3 × 2 × 1 × 3 (do first) = 18
8. Must be greater than 5 000.       ANS: 2 × 3 × 2 × 1 = 12
9. Must be greater than 400.      ANS: 3 × 3 × 2 + 4 × 3 × 2 × 1 = 18 + 24 = 42
10. 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

1. How many ways can 6 books be arranged on a shelf?

ANS: 6P6 or 6 × 5 × 4 × 3 × 2 × 1 = 720

1. 5-letter arrangements are formed from the letters in NUMBERS.  How many if:
2.     no vowels?    ANS: 5 × 4 × 3 × 2 × 1 = 120
3.     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

1. 6-letter arrangements are formed from ABCDEF.  How many if:
1. no restrictions?

ANS: 6 × 5 × 4 × 3 × 2 × 1 = 6! = 720

1.     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.

1.     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.

1.     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.

1.     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.

1.     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

1. How many arrangements of the letters in the word:
1. BANANA

ANS:     3 A’s and 2 N’s

1. 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:

1.     no restrictions?        ANS: 12C3 = 220
2.     all boys?        ANS: 7C3 = 35
3.     1 girl and 2 boys?        ANS: 5C1 ∙ 7C2 = 105
4.     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

1. 10 points are on a circle (no 3 are collinear).  How many:
1. 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.