General Statistics<\/p>\n
Name<\/p>\n
Affiliation<\/p>\n
Choosing an Outfit<\/p>\n
3 pairs of pants (B, R, O)<\/p>\n
2 tops (W, R)<\/p>\n
2 shoes (B, W)<\/p>\n
a.Tree Diagram<\/p>\n
# of outfits = 3 \u00b4 2 \u00b4 2 = 12<\/p>\n
2.How many 3-digit numbers can be formed if:<\/p>\n
a. there are no restrictions? ANS: 9 \u00b4 10 \u00b4 10 = 900 (a 3 digit number can\u2019t start with zero)<\/p>\n
b.there are no repetitions? ANS: 9 \u00b4 9 \u00b4 8 = 648 (gain the zero, lose the 1st digit number)<\/p>\n
3.How many 4-letter arrangements can be formed if:(26 letters in the alphabet?)<\/p>\n
a.there are no repetitions? ANS: 26 \u00b4 25 \u00b4 24 \u00b4 23 = 358 800<\/p>\n
b.repetitions allowed? ANS: 264 = 456 976 a.k.a. 26 x 26 x 26 x 26 =<\/p>\n
4.How many numbers can be formed from the digits 1, 4, 6, and 8 (one of each) if the number:<\/p>\n
a. Must be even ANS: 3 \u00b4 2 \u00b4 1 \u00b4 3 (do first) = 18<\/p>\n
b. Must be greater than 5 000. ANS: 2 \u00b4 3 \u00b4 2 \u00b4 1 = 12<\/p>\n
C. Must be greater than 400. ANS: 3 \u00b4 3 \u00b4 2 + 4 \u00b4 3 \u00b4 2 \u00b4 1 = 18 + 24 = 42<\/p>\n
1.How many 5-letter arrangements can be made from the letters A, B, C, D, E, F, G? <\/p>\n
ANS: 7P5 or 7 \u00b4 6 \u00b4 5 \u00b4 4 \u00b4 3 = 2 520<\/p>\n
2.How many ways can 6 books be arranged on a shelf? <\/p>\n
ANS: 6P6 or 6 \u00b4 5 \u00b4 4 \u00b4 3 \u00b4 2 \u00b4 1 = 720<\/p>\n
3.5-letter arrangements are formed from the letters in NUMBERS. How many if:<\/p>\n
a.no vowels? ANS: 5 \u00b4 4 \u00b4 3 \u00b4 2 \u00b4 1 = 120<\/p>\n
b.1st and last letters are consonants? ANS: 5 \u00b4 5 \u00b4 4 \u00b4 3 \u00b4 4 = 1 200<\/p>\n
___ x ___ x ___ x ___ x ____AEIOU are vowels in the alphabet<\/p>\n
Con Any Any Any Con<\/p>\n
4. 6-letter arrangements are formed from ABCDEF. How many if:<\/p>\n
no restrictions? <\/p>\n
ANS: 6 \u00b4 5 \u00b4 4 \u00b4 3 \u00b4 2 \u00b4 1 = 6! = 720<\/p>\n
b.A & B together? (think of AB as one letter) <\/p>\n
ANS: 5 \u00b4 4 \u00b4 3 \u00b4 2 \u00b4 1 \u00b4 2 \u00b4 1 = 120<\/p>\n
Number of ways to arrange AB \u00bf<\/p>\n
The 2 \u00b4 1 is the placing of the A and B<\/p>\n
The 5 \u00b4 4 \u00b4 3 \u00b4 2 \u00b4 1 are the placing of the AB as one unit and the CDEF as 4 units.<\/p>\n
c.A, B, & C together? ANS: 4 \u00b4 3 \u00b4 2 \u00b4 1 \u00b4 3 \u00b4 2 \u00b4 1 = 144<\/p>\n
The 4 \u00b4 3 \u00b4 2 \u00b4 1 is the arranging of DEF (in any order) and the ABC as a whole.<\/p>\n
The 3 \u00b4 2 \u00b4 1 is the arranging of the ABC.<\/p>\n
d.A & B separated? (arrange others first) ANS: 4 \u00b4 3 \u00b4 2 \u00b4 1 \u00b4 5 \u00b4 4 = 480 OR #a – #b<\/p>\n
The 4 \u00b4 3 \u00b4 2 \u00b4 1 is the CDEF <\/p>\n
The 5 is the A or B (one or the other) and the CDEF<\/p>\n
You basically have __ C __ D __ E __ F ___<\/p>\n
Now the CDEF can go in any order hence the 4!<\/p>\n
The A or B could go in any of the 5 spaces hence the multiplying 5<\/p>\n
Once the A or B is placed, the other one has 4 difference places to go, hence the multiplying 4.<\/p>\n
e.A, B, & C separated? ANS: 3 \u00b4 2 \u00b4 1 \u00b4 4 \u00b4 3 \u00b4 2 = 144 (can’t do as above)<\/p>\n
Same hint as above. You basically have __ D __ E __ F ___<\/p>\n
Now the DEF can go in any order hence the 4!<\/p>\n
The A or B or C could go in any of the 4 spaces hence the multiplying 4<\/p>\n
Once the A B or C is placed, the other two have 3 difference places to go, hence the multiplying 3.<\/p>\n
The multiplying 2 is the last A, B or C to go in either of the empty two spots.<\/p>\n
f.the word BAD appears? ANS: 4 \u00b4 3 \u00b4 2 \u00b4 1 = 24<\/p>\n
The 4 is the word BAD or C, E, or F (4 items as BAD is one unit)<\/p>\n
The 3 is the 4 options above less the one already used.<\/p>\n
The 2 is the 3 options above less the one already used.<\/p>\n
The 1 is the only option left.<\/p>\n
BAD cannot be switched as it must remain as BAD<\/p>\n
5. How many arrangements of the letters in the word:<\/p>\n
BANANA <\/p>\n
ANS: EMBED Equation.DSMT4 3 A\u2019s and 2 N\u2019s<\/p>\n
STATISTICS <\/p>\n
ANS: EMBED Equation.DSMT4 3 S\u2019s, 3 T\u2019s and 2 I\u2019s<\/p>\n
Computation<\/p>\n
A committee of 3 is selected from 5 girls and 7 boys. How many if:<\/p>\n
a.no restrictions? ANS: 12C3 = 220<\/p>\n
b.all boys? ANS: 7C3 = 35<\/p>\n
c.1 girl and 2 boys? ANS: 5C1 \u00b7 7C2 = 105<\/p>\n
d.At least 1 girl? ANS: 1 girl or 2 girls or 3 girls = 5C1 \u00b7 7C2 + 5C2 \u00b7 7C1 + 5C3<\/p>\n
= 105 + 70 + 10 <\/p>\n
= 185<\/p>\n
OR #a – #b = 220 – 35<\/p>\n
= 185<\/p>\n
2. 10 points are on a circle (no 3 are collinear). How many:<\/p>\n
Different lines are determined? <\/p>\n
ANS: 10C2 = 45<\/p>\n
781685152400A line is a connection of 2 points. So, take the 10 points and connect 2 of them.<\/p>\n
quadrilaterals? <\/p>\n
ANS: 10C4 = 210<\/p>\n
A quadrilateral has 4 sides. So, take 10 points and connect 4 of them.<\/p>\n","protected":false},"excerpt":{"rendered":"
General Statistics Name Affiliation Choosing an Outfit 3 pairs of pants (B, R, O) 2 tops (W, R) 2 shoes<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[],"tags":[],"class_list":["post-40063","post","type-post","status-publish","format-standard","hentry"],"yoast_head":"\n