I wish to know the difference between permutation and combination

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Posted on Feb 01, 2009

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Can you help me solve the permutation and combination program in c++???

Posted on Sep 25, 2008

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Permutation : Permutation means *arrangement* of things. The word *arrangement *is used, if the order of things *is considered*.

Combination: Combination means *selection* of things. The word *selection* is used, when the order of things has *no importance*.

Posted on Jul 11, 2008

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Posted on Jan 02, 2017

If the order doesn't matter there is one combination of five numbers, namely 13457.

If the order matters then there are 120 permutations of the five digits.

If the order matters then there are 120 permutations of the five digits.

Jan 27, 2015 | Home Security

1 to 9 sounds wrong, it should ben 10 to 9 if it is permutation order 10

Aug 13, 2008 | Microsoft Computers & Internet

Do you want combinations or permutations? They're different things. A combination of n items taken m at a time, sometimes written "nCm", selects m items from a group of n without regard to order. A permutation of n items taken m at a time, sometimes written "nPm", selects m items from a group of m with regard to order.

Given six numbers, there are six combinations of one number, 15 combinations of two numbers, 20 combinations of three numbers, 15 combinations of four numbers, six combinations of five numbers, and one combination of six numbers.

Given six numbers, there are six permutations of one number, 30 permutations of two numbers, 120 permutations of three numbers, 360 permutations of four numbers, 720 permutations of five numbers, and 720 permutations of six numbers.

Given six numbers, there are six combinations of one number, 15 combinations of two numbers, 20 combinations of three numbers, 15 combinations of four numbers, six combinations of five numbers, and one combination of six numbers.

Given six numbers, there are six permutations of one number, 30 permutations of two numbers, 120 permutations of three numbers, 360 permutations of four numbers, 720 permutations of five numbers, and 720 permutations of six numbers.

Oct 22, 2014 | Office Equipment & Supplies

There is only one combination using all five digits, namely 1, 2, 3, 4, 7 in some order.

If the order matters then there are 120 different permutations. I won't list them all here, but they start with 1, 2, 3, 4, 7 and go through 7, 4, 3, 2, 1.

If the order matters then there are 120 different permutations. I won't list them all here, but they start with 1, 2, 3, 4, 7 and go through 7, 4, 3, 2, 1.

Aug 21, 2014 | Computers & Internet

Does every number have to be taken only once?

If order is important you have 11P7=1663200 permutations (different from combinations)

If order is not important for example 1,2,3,4,5,6,7---- 2,1,3,4,5,6,7----2,3,1,4,5,6,7----2,3,4,1,5,6,7 etc are all one and the same then you have

11C7 =330 combinations.

You get them by mere obstinacy: take a pad and start. It is only 330.

If order is important you have 11P7=1663200 permutations (different from combinations)

If order is not important for example 1,2,3,4,5,6,7---- 2,1,3,4,5,6,7----2,3,1,4,5,6,7----2,3,4,1,5,6,7 etc are all one and the same then you have

11C7 =330 combinations.

You get them by mere obstinacy: take a pad and start. It is only 330.

Dec 13, 2013 | Office Equipment & Supplies

That depends on how many of those six numbers you take.

If you only take one number, there are six combinations: {1}, {2}, {3}, {4}, {5}, and {6}.

If you take two numbers, there are fifteen combinations: {1,2}, {1,3}, {1,4}, {1,5}, {1,6}, {2,3}, {2,4}, {2,5}, {2,6}, {3,4}, {3,5}, {3,6}, {4,5}, {4,6}, and {5,6}.

If you take three numbers, there are twenty combinations.

If you take four numbers, there are fifteen combinations.

If you take five numbers, there are six combinations.

If you take all six numbers, there is only one combination: {1,2,3,4,5,6}.

In general, if you take 'm' objects out of a set of 'n' objects, the number of combinations is given by n!/[(m!)(n-m)!] where '!' is the factorial operator.

If you only take one number, there are six combinations: {1}, {2}, {3}, {4}, {5}, and {6}.

If you take two numbers, there are fifteen combinations: {1,2}, {1,3}, {1,4}, {1,5}, {1,6}, {2,3}, {2,4}, {2,5}, {2,6}, {3,4}, {3,5}, {3,6}, {4,5}, {4,6}, and {5,6}.

If you take three numbers, there are twenty combinations.

If you take four numbers, there are fifteen combinations.

If you take five numbers, there are six combinations.

If you take all six numbers, there is only one combination: {1,2,3,4,5,6}.

In general, if you take 'm' objects out of a set of 'n' objects, the number of combinations is given by n!/[(m!)(n-m)!] where '!' is the factorial operator.

Apr 30, 2013 | Mathsoft Computers & Internet

I must know which calculator you speak of. They're all a little different. If you speak of Windows calculator, get this calculator: http://www.calculator.org/default.aspx

Apr 22, 2010 | Computers & Internet

If you have an older version on the TI30x Solar (not TI30Xa or newer) then you must use a different key sequence for nPr/nCr.

52 [2nd][x/y] 5 [2nd][nCr] gives 2598960 (the x/y button is above the Pi symbol on my version).

8 [2nd][x/y] 3 [2nd][nPr] gives 336

This took me forever to find out, so hopefully it will help someone!

52 [2nd][x/y] 5 [2nd][nCr] gives 2598960 (the x/y button is above the Pi symbol on my version).

8 [2nd][x/y] 3 [2nd][nPr] gives 336

This took me forever to find out, so hopefully it will help someone!

Jan 27, 2010 | Texas Instruments TI-84 Plus Silver...

Hello,

Use the sequence [2ndF][nPr] for permutations and [2ndF][nCr] for combinations.

Exemples: 10(nPr)3 is entered as 10[2ndF][nPr]3 [=] gives 720

5(nCr)2 is entered as 5 [2ndF][nCr] 2 [=] gives 10

Hope it helps.

Use the sequence [2ndF][nPr] for permutations and [2ndF][nCr] for combinations.

Exemples: 10(nPr)3 is entered as 10[2ndF][nPr]3 [=] gives 720

5(nCr)2 is entered as 5 [2ndF][nCr] 2 [=] gives 10

Hope it helps.

Apr 09, 2009 | Sharp EL-531VB Calculator

You can enter combinations and permutations.

If the order of the selection matters (ie. 1st, 2nd, and 3rd place out of 40 contestants) use permutation. Enter:

40{shift}{nPr}3 and then [=] for your answer set (=59280 possible ways).

IF order doesn not matter (ie. a team of 3 from 40 possible members) then it is a combination. Do the same as above, but substitute {nCr} for {nPr}. (= 9880 possible ways).

For your question: a three-digit set, selecting from 0-9. There is no specified order, it is a combination. The total n=10 (there are 10 digits in the range 0-9) and you are selecting 3.

so.... 10[nCr]3 with give the number of possible combinations.

If the order of the selection matters (ie. 1st, 2nd, and 3rd place out of 40 contestants) use permutation. Enter:

40{shift}{nPr}3 and then [=] for your answer set (=59280 possible ways).

IF order doesn not matter (ie. a team of 3 from 40 possible members) then it is a combination. Do the same as above, but substitute {nCr} for {nPr}. (= 9880 possible ways).

For your question: a three-digit set, selecting from 0-9. There is no specified order, it is a combination. The total n=10 (there are 10 digits in the range 0-9) and you are selecting 3.

so.... 10[nCr]3 with give the number of possible combinations.

Feb 16, 2009 | Casio FX-115ES Scientific Calculator

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