# how to find the number of arrangements possible for a particular draw from crads

• Jul 12th 2010, 01:44 AM
dijinj
how to find the number of arrangements possible for a particular draw from crads
There are 25 cards with 5 signs and each sign contains digits 1 to 5 in them. What is the maximum number of arrangements possible in 5 cards drawn from them? numbers can be repeated but signs cannot be repeated and in each draw of 5 cards there should be all the five signs in it without repetition and without the absence of any sign. What is the general formula for this? Say n signs X n numbers = n^2 cards
• Jul 12th 2010, 02:50 AM
Hello dijinj

I'm sorry, I just don't understand what the question means?
Quote:

Originally Posted by dijinj
There are 25 cards with 5 signs and each sign contains digits 1 to 5 in them...

So a 'sign' contains digits 1 to 5. How many digits make up a sign? All of them? Just one? What?
Quote:

...numbers can be repeated but signs cannot be repeated...
What is meant by a 'number'? A set of digits? A single digit? What?

Unless you can define much more accurately what you mean, it's not going to be possible to give you any help.

Do you have a statement of the original question? If so, let's see it in full, and we'll take it from there.

• Jul 12th 2010, 05:22 AM
dijinj
just like normal playing cards has 52 cards with four signs( hearts, ispade, etc) and each sign( for exapmple ispade) has a number among 1-10 then j k q. this system has 25 cards and five signs each sign has a number among 1-5 on them. total no: of cards bearing same sign is five just like in playing cards, where it is 13.

In a draw the numbers can be repeated for example in one draw all the five card can be of number 5 or four cards can be 5 and fifth one can be say 2. but in a draw the sign can not be repeated it must be unique for each cards that has been drawn. for example first card can be claver second card can be dice,, third can be hearts forth can be ispade and fifth can be with a circle sign on it, order of signs is not important but it must be unique in each card in every draw.
• Jul 12th 2010, 05:40 AM
dijinj
Can permutations with repetition formula can directly be applied to here (n^r), problem is uniqueness of signs. only number on cards can be repeated
• Jul 12th 2010, 06:10 AM
Hello dijinj

OK. So let me see if I understand the question. A pack of 25 cards contains five 'suits' (signs), each 'suit' containing five different cards. We simply have to find the number of possible arrangements of five cards, where each card is chosen from a different 'suit'.

There are $\displaystyle \displaystyle 5!$ ways of arranging the different 'suits'. There are $\displaystyle \displaystyle 5$ ways of choosing each card from within its suit. So the total number of arrangements is $\displaystyle \displaystyle 5^5\times5!$.
If you didn't really mean the number of different arrangements, but instead you meant the number of different selections, then the answer is just $\displaystyle \displaystyle 5^5$.

• Jul 12th 2010, 06:57 AM
Soroban
Hello, dijinj!

I must make some assumptions . . .

Quote:

There are 25 cards with 5 suits and each suit contains digits 1 to 5.

What is the maximum number of arrangements of 5 cards drawn from them?
All five suits must be in the arrangement.

It says "arrangements"; I assume that the order of the cards is important.

We have these 25 cards:

. . $\displaystyle \begin{array}{ccccc}A\spadesuit & 2\spadesuit & 3\spadesuit & 4\spadesuit & 5\spadesuit \\ A\heartsuit & 2\heartsuit & 3\heartsuit & 4\heartsuit & 5\heartsuit \\ A\clubsuit & 2\clubsuit & 3\clubsuit & 4\clubsuit & 5\clubsuit \\ A\diamondsuit & 2\diamondsuit & 3\diamondsuit & 4\diamondsuit & 5\diamondsuit \\ A\bigstar & 2\bigstar & 3\bigstar & 4\bigstar & 5\bigstar \end{array}$

The first card can be any of the 25 cards.

The second card can be any of the 20 cards of a second suit.

The third card can be any of the 15 cards of a third suit.

The fourth card can be any of the 10 cards of a fourth suit.

The fifth card must be one the remaining 5 cards of the fifth suit.

There are: .$\displaystyle 25\cdot20\cdot15\cdot10\cdot5 \:=\:375,000$ possible arrangements.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

If the order of the cards is not important, divide by $\displaystyle 5!$

. . $\displaystyle \text{There are: }\;\dfrac{375,000}{120} \;=\;3125\:\:selections \text{ of 5 cards.}$

• Jul 12th 2010, 11:58 PM
dijinj