1. ## Discrete Maths

Hello again =)
So lets see !

1) We 've got a pouch with M&Ms candies . There are 4 Blue , 7 Orange , 8 Pink and 5 Yellow. Nick puts his hand into the pouch, and takes 6 candies.How many ways there exist , so that into the pouch there exist at least one Blue candy?

I believe, that 1st we must find out how many different combinations of 6 candies can be done. I think its 24^6 = 191,102,976. Then i think that we have to find out how many combinations of 6 candies that include 4xBlues and 2 others can be done, and just subtract it.

Basically i think the solution is 24^6 - 20^2 = 191,102,576

2) How many 6-digit numbers can be made if:

• The First digit must be 6,7 or 8
• The last digit must be < = 5
• Every digit must be unique

Its 3x10x10x10x10x6 =180,000 . But, i think that, in this way, its not every digit unique. Any suggestions?

Thanks in advance for any responses =)

2. ## n choose k

1) For this question, the number of ways of picking 6 candies out of 24 is not $\displaystyle 24^{6},$ that would be the number different sequences you can get by picking six 6 but replacing each one as you put it back. The number of ways of choosing $\displaystyle k$ candies from a bag of $\displaystyle n$ can be shown to be
$\displaystyle \frac{n!}{(n-k)!k!}.$
You can find an explanation of this here: Combination - Wikipedia, the free encyclopedia

3. Hello, primeimplicant!

2) How many 6-digit numbers can be made if:

. . The first digit must be 6, 7 or 8
. . The last digit must be < 5
. . Every digit must be unique

The first digit is from: $\displaystyle \{6, 7, 8\}\;\;\hdots\;\text{3 choices.}$

The last digit is from: $\displaystyle \{0, 1, 2, 3, 4, 5\}\;\;\hdots\;\text{6 choices.}$

The middle 4 digits are selected from the remaining 8 digits . . . $\displaystyle _8P_4 = 1680\text{ choices.}$

Answer: .$\displaystyle 3 \times 6 \times 1680 \;=\;30,240$

4. Thanks, you make it soooo easy! Artist!

5. Regarding my 1st question,
with nimon's help, i believe its like this:

$\displaystyle \binom{24}{6}-\binom{20}{2}= 134,596 - 190= 134,406$

So i believe 134,406 is the final answer , correct me if i 'am wrong =)

6. Originally Posted by primeimplicant
Regarding my 1st question,
with nimon's help, i believe its like this:

$\displaystyle \binom{24}{6}-\binom{20}{2}= 134,596 - 190= 134,406$

So i believe 134,406 is the final answer , correct me if i 'am wrong =)
At least one blue: $\displaystyle \binom{24}{6}-\binom{20}{6}$

7. $\displaystyle \binom{24}{6}-\binom{20}{6}$

$\displaystyle =(number\ of\ ways\ to\ choose\ any\ 6)-(number\ of\ ways\ to\ choose\ zero\ blues)$

is the number of ways to choose at least 1 blue.

As this includes choosing 4 blues, there will be no blues left in the pouch for many of these selections.

$\displaystyle \binom{24}{6}-\binom{20}{2}$

is correct.

$\displaystyle \binom{24}{6}-\binom{20}{2}=\binom{4}{0}\binom{20}{6}+\binom{4}{ 1}\binom{20}{5}+\binom{4}{2}\binom{20}{4}+\binom{4 }{3}\binom{20}{3}$

8. alright, i feel i little bit lost, and dizzy.

Again , i want to find out how many ways there exist so that into the pouch, theres at least 1 Blue candy, Not into Nick's hand .

So its

$\displaystyle \binom{24}{6}-\binom{20}{2}$ As i said before right?

Edit: $\displaystyle \binom{24}{6}-\binom{20}{2}=\binom{4}{0}\binom{20}{6}+\binom{4}{1}\binom{20}{ 5}+\binom{4}{2}\binom{20}{4}+\binom{4}{3}\binom{20 }{3}$

I Dont understand the Bold part, i would i appreciate if you could link me some wiki related theory or something.

Thanks again

Edit2: ok nvm found it

9. That's it, primeimplicant,

We have to distinguish between
"at least one in the hand", or
"at least one in the pouch".

You're correct.

10. $\displaystyle \binom{24}{6}-\binom{4}{4}\binom{20}{2}=\binom{4}{0}\binom{20}{6 }+\binom{4}{1}\binom{20}{5}+\binom{4}{2}\binom{20} {4}+\binom{4}{3}\binom{20}{3}$

This is

(select any 6 from the 24)-(the number of ways to pair all 4 blues with 2 others from the remaining 20)

and this will give us the number of ways to have at least 1 blue remaining in the pouch.

This is the same as calculating the removal from the pouch of

No blues with any 6 from the remaining 20
+ any blue from the 4 blues matched with 5 other non-blues
+ any 2 blues from the 4 with any 4 non-blues from the 20 non-blues
+ any 3 blues with any 3 non-blues

as we must avoid taking out all 4 blues.