A few different statistics questions - Finite Math

1.) There are 4 women and 8 men eligible for awards. No person can get two awards. In how many ways can two identical awards of $300 and two identical awards of $500 be given?


2.) A professor assigns a 4-digit code to each of his students for the purpose of posting grades.
a) How many different codes are possible?
b) How many codes contain no fives?
c) How many codes contain exactly one five?
d) How many codes contain exactly two fives



3.) A bag contains 5 red marbles, 4 blue marbles, 3 green marbles, and 2 yellow marbles. Victor
reaches into the bag and grabs 3 marbles at random.
a) How many groups of three marbles contain at least one yellow marble?
b) How many groups of three marbles contain at most one yellow marble?


4.) There are 10 students in a drama class, and a group of 4 students will be selected to attend a
theater performance.
a) How many different ways are there to select the group of 4 students?
b) Suppose that Gabriella is one of the 10 students in the drama class. How many
4-person groups are there that contain Gabriella?

5.) Your debit card requires you to select a 6 digit PIN (a sequence of 6 digits, each from 0 – 9). You always select a PIN with exactly 3 threes, 2 twos, and 1 one. Example: 233123. How
many different ways are there for you to select your PIN under these conditions?




If anyone could even help me with even one of those I would appreciate it. Thanks!

Hello, Vitality!

2.) A professor assigns a 4-digit code to each of his students for the purpose of posting grades.

a) How many different codes are possible?

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For each of the four digits, there are 10 choices.
. . There are: .\(\displaystyle 10^4 \,=\,10,\!000\) possible codes.

b) How many codes contain no fives?

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For each of the four digits, there are 9 choices.
. . There are: .\(\displaystyle 9^4 \,=\,6,\!561\) possible codes with no 5's.

c) How many codes contain exactly one Five?

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The code contains a Five. .It can occupy any of 4 positions.
The other 3 digits have 9 choices each: .\(\displaystyle 9^3\,=\,729\) choices.

There are: .\(\displaystyle 4 \times 729 \:=\:2,\!916\) codes with one Five.

d) How many codes contain exactly two Fives?

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The code contains two Fives. .They can occupy any of \(\displaystyle _4C_2 \,=\,6\) positions.
The other two digits have 9 choices each: .\(\displaystyle 9^2 \,=\,81\) choices.

There are: .\(\displaystyle 6 \times 81 \:=\:486\) codes with two Fives.

Hello again, Vitality!

3.) A bag contains 5 red, 4 blue, 3 green, and 2 yellow marbles.
Victor reaches into the bag and grabs a sample of 3 marbles at random.

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There are 14 marbles; a sample of 3 marbles is drawn.

There are: .\(\displaystyle _{14}C_3\:=\:\frac{(14)(13)(12)}{(3)(2)(1)} \:=\:364\) possible samples.

a) How many samples contain at least one yellow marble?

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The opposite of "at least one yellow" is "no yellow".

There are: .\(\displaystyle _{12}C_3 \:=\:\frac{(12)(11)(10)}{(3)(2)(1)} \:=\:220\) samples with no yellow marbles.

Therefore, there are: .\(\displaystyle 364 - 220 \:=\:144\) samples with at least one yellow marble.

b) How many samples contain at most one yellow marble?

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"At most one yellow" means "one yellow" or "no yellow".

One yellow (and two Others): .\(\displaystyle (_2C_1)(_{12}C_2) \:=\2)(66) \:=\:132\) samples.

From (a), no yellow: .\(\displaystyle {\bf220}\) samples.

Therefore, there are: .\(\displaystyle 132 + {\bf220} \:=\:{\bf352}\) samples with at most one yellow marble.

Thanks very much for the explanation! Just one little correction in one of your answers.

soroban said:

The opposite of "at least one yellow" is "no yellow".

There are: .\(\displaystyle _{12}C_3 \:=\:\frac{(12)(11)(10)}{(3)(2)(1)} \:=\:220\) samples with no yellow marbles.

Therefore, there are: .\(\displaystyle 364 - 220 \:=\:144\) samples with at least one yellow marble.




"At most one yellow" means "one yellow" or "no yellow".

One yellow (and two Others): .\(\displaystyle (_2C_1)(_{12}C_2) \:=\2)(66) \:=\:132\) samples.

From (a), no yellow: .\(\displaystyle 144\) samples.

Therefore, there are: .\(\displaystyle 132 + 144 \:=\:276\) samples with at most one yellow marble.

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Your wrote: 132 + 144 = 276
It should be: 132 + 220 = 352

I'm sure you just looked at the wrong number

Hello, Vitality!

1) There are 4 women and 8 men eligible for awards. .No person can get two awards.
In how many ways can two identical awards of $300 and two identical awards of $500 be given?

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From the 12 people, choose 2 to receive the $300 awards.
[COLOR=#be0be]. . [/COLOR]There are: .\(\displaystyle _{12}C_2 \,=\,66\) ways.

From the 10 remaining people, choose 2 to receive the $500 awards.
. . There are: .\(\displaystyle _{10}C_2 \,=\,45\) ways.

Therefore, there are: .\(\displaystyle 66\times45 \:=\:2970\) ways to distribute the awards.

4) There are 10 students in a drama class.
A group of 4 students will be selected to attend a theater performance.

a) How many different ways are there to select the group of 4 students?

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There are: .\(\displaystyle _{10}C_4 \:=\:210\) possible groups.

b) Suppose that Gabriella is one of the ten students in the drama class.
How many 4-person groups are there that contain Gabriella?

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Gabriella is in the group.
From the remaining 9 students, choose 3 more students.

. . There are: .\(\displaystyle _9C_3 \:=\:84\) groups containing Gabriella.

5) Your debit card requires you to select a 6 digit PIN (a sequence of 6 digits, each from 0 – 9).
You always select a PIN with exactly three 3's, two 2's, and one 1. .Example: 233123.
How many different ways are there for you to select your PIN under these conditions?

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If you used six different digits, there would be: .\(\displaystyle 6!\) possible PINs.

But there are three identical 3's and two identical 2's.

Hence, there are: .\(\displaystyle \dfrac{6!}{3!\,2!} \:=\:60\) possible PINs.

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An alternate solution ...

The PIN is a 6-digit number: . _ _ _ _ _ _

Choose 3 of the 6 positions for the 3's.
. . There are: .\(\displaystyle _6C_3\,=\,20\) choices.

Choose 2 of the remaining 3 positions for the2's.
. . There are: .\(\displaystyle _3C_2 \,=\,3\) choices.

There is one position remaining for the 1.


Therefore, there are: .\(\displaystyle 20\cdot3\cdot1 \:=\:60\) possible PINs.

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A trivial observation . . .

Neither of us used the term "PIN number".

Obviously, "Personal Identification Number number" is redundant.

The same is true for "GPS system" and "ATM machine".