1 combination symbols a supplement to greenleafs qr text compiled by samuel marateck ©2009
TRANSCRIPT
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1
Combination Symbols
A supplement to Greenleaf’s QR Text
Compiled by Samuel Marateck ©2009
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2
How many 4-card hands consisting of
1 king and 3 queens can be chosen
from a deck?
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How many 4-card hands consisting of1 king and 3 queens can be chosenfrom a deck?
Since order does not matter and there arefour kings and four queens in the deck,the answer is:
( 4 1) ( 4
3)
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4
What is the meaning of ( 4 1)?
It’s the number of ways we can choose one
thing from four, independent of the order.
It is pronounced “four choose one”.
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Similarly ( 4 3) is the number of ways we can
choose three things from four independent
of the order. It is pronounced “four choose
three”.
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In ( 4 1) ( 4
3), why do we multiply the
two?
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For each king there are three queen pairings.
These are the pairings for the king of spades:
k♠ Q♠ Q♣ Q♥
k♠ Q♠ Q♣ Q♦
k♠ Q♣ Q♥ Q♦
k♠ Q♥ Q♦ Q ♠
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But there are also k♥, k♣ and k♦. So there
are 16 different combinations, four for each
King.
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What is the probability of choosing
4-card hands consisting of 1 king and 3
queens from a deck?
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10
What is the probability of choosing
4-card hands consisting of 1 king and 3
queens from a deck?
( 4 1) ( 4
3) / ( 52 4)
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We divide by ( 52 4) since this is the number
of ways we can choose four cards at
random from a deck.
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Let’s evaluate ( 4 1) ( 4
3) / ( 52 4)
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( 4 1) ( 4
3) / ( 52 4) is:
16/(52*51*50*49/(4*3*2*1))
=0.00006 or .006%
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Out of how many hands would you expect
to get this hand?
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Out of how many hands would you expect
to get this hand?
0.00006 is 6x 10-5 , so in 105 hands you
would expect to get 6 such hands or
in one out of 16,666 hands you would get
this hand.
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How many 5-card hands can you get that
have three aces?
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How many 5-card hands can you get that
have three aces?
The number of ways we can choose three
aces is ( 4 3) . How many cards are left in
the deck?
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How many non-aces are in the deck?
There are 48 non-aces left in the deck and
there are two more cards to choose for our
hand.
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So there are ( 4 3) ( 48
2) ways we can get
three aces:
4*48*47/2 = 4*47*24 = 4512 ways.
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What is the probability of getting three
aces in a 5-card hand?
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What is the probability of getting three
aces in a 5-card hand?
( 4 3) ( 48
2) / ( 52 5) =
4512/((52*51*50*49*48)/(5*4*3*2*1)) =
4512/2598960 = .00174
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What is the probability of winning the
lottery?
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What is the probability of winning the
lottery?
There are 54 numbers that you can choose
from; the numbers 1 to 54. You must choose
the five correct numbers independent of
their order. The answer is:
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P(winning) = 1/( 54 5)
( 54 5) = 54*53*52*51*50/120
1/( 54 5) = 3.16 x 10-7
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25
If there are 6 pegs distributed in a circle and
a line is drawn from each peg to each other
peg, how many lines are there?
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For each peg 5 lines are drawn; but there
are 6 pegs. Since, however, each line
connects two pegs, we are overcounting
by 2, so we must divide by 2.
What is the answer?
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# of lines is 5*6/2 or 15.
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Another way of looking at this is:
From the first peg, 5 lines are drawn. From
the second peg, 4 lines are drawn since it
is already connected to the first peg. From
the third peg, 3 lines are drawn, since it is
connected to the first two, and so on,
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For the six pegs, 5+4+3+2+1 or 15 lines are
drawn. For n pegs n-1 + n-2 + n-3 +..+ 1
lines are drawn. We know what the sum
from 1 to m is.
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The sum is: m(m+1)/2.
Substituting n-1 for m, the sum from 1 to
n-1 is (n-1)(n-1 +1)/2 =?
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(n-1)(n-1 +1)/2 = n(n-1)/2 which is the
answer we got before.
Can we do this with combination symbols?
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32
If there are 6 pegs distributed in a circle and
a line is drawn from each peg to each other
peg, how many lines are there?
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There are 6 slots:
. . . . . ..
1 2 3 4 5 6
How many ways can we place two item
in these slots?
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How many ways can we place two item
in these slots?
The answer is ( 6 2).
For n pegs it’s ?
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For n pegs it’s ( n 2).
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How many ways can we choose a 5-card
hand so that no two cards have the same
face values?
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37
How many ways can we choose a 5-card
hand so that no two cards have the same
face values?
For the first card we have ( 52 1) ways we
can choose the first card. How many
choices do we have for the second card?
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How many choices do we have for the
second card?
48, since one face value has been
eliminated. So the number of ways we can
choose the second card is:
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( 48 1).
The third card is?
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( 44 1).
So the final answer is:
( 52 1) ( 48
1)( 44 1) ( 40
1) ( 36 1).
What is the probability?
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P(each card has a different face value) =
( 52 1) ( 48
1)( 44 1) ( 40
1) ( 36 1)
( 52 5)
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42
In a class of 25, what is the probability that
two or more people have the same
birthdate?
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43
In a class of 25, what is the probability that
two or more people have the same
birthdate?
We will first calculate the probability that no
one has the same birthdate.
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Given the first person, the probability that the second one has a different birth date is364/365. That the first, second and third ones have different birth dates is:1* 364/365*363/365.
For all 25 people?
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For all 25 people?
P(different birth dates) =
364*363*362*361…341/36524 = 0.47
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46
P(2 or more have same birth dates) = .53
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There are 25 people to be chosen for a
Committee or 5. What is my probability of
being chosen?
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What is the probability of my being chosen?
( 1 1) ( 24
4)/ ( 25 5).
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49
An urn contains 10 red balls and 40 black
ones. What is the probability you will draw
2 red balls.
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( 10 2) ( 40
0)/ ( 50 2) = 10*9/2 /(50*49/2)
= 45/1225
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51
An urn contains 17 red balls and 33 black
ones. What is the probability you will draw
7 red balls if you choose 10 randomly?
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( 17 7) ( 33
3)/ ( 50 10)
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53
A jury pool contains 98 men and 75 women.
12 jurors are chosen at random. What is
the probability that 6 will be women
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54
( 98 6) ( 75
6)/ ( 173 12)