A bag has 4 marbles purple, green, blue and red. Adam pulls out 3 marbles w/o looking. How many different combinations of 3 colors are possible?(note 4th grade math no complicated formulas)  The...

A bag has 4 marbles purple, green, blue and red. Adam pulls out 3 marbles w/o looking. How many different combinations of 3 colors are possible?

(note 4th grade math no complicated formulas)  The one thing that is confusion is i do not know if red, blue, purple is the same as purple, blue, red..I would assume it is the same and order does not matter

pohnpei397 | College Teacher | (Level 3) Distinguished Educator

Posted on

What you are doing here is (as your tag says) combinations.  This is as opposed to permutations.  The difference is that, with combinations, order doesn't matter (Adam is grabbing them all at once, not one at a time).  Permutations are when he grabs them one at a time.

The correct answer for your question is four -- there should be four different combinations of three marbles that Adam could grab.  The possibilities are

• purple, blue, green
• purple, green, red
• purple, blue, red
• blue, red, green

Any other combination would be a repeat (like blue, red, purple), just in a different order.

krishna-agrawala | College Teacher | (Level 3) Valedictorian

Posted on

When the order in which the marbles of different colours are pulled out is not considered, then the question can be answered very simply by modifying it the different possibilities regarding the fourth marble that is left in the bag after three marbles have been pulled out of the bag. This is because just by looking at the the fourth marble, we can tell which other three marbles have been pulled out.

There are just 4 different possibilities of the fourth marble left in the bag - 1) purple, 2) green, 3) blue and 4) red. Corresponding to this there are four combinations of three marbles pulled out. These are:

1) green, blue and red,

2) purple, blue and red,

3) purple, green and red,

4) purple, green, and blue.

neela | High School Teacher | (Level 3) Valedictorian

Posted on

In combinations , the order is to be ignored. In permutations orders or arrangements are to be considered.

From four marbles , green, blue,red and adam, how many combunatuions are possible? The number of combinations of three is as mmany as how many possibility of remaining one you leave when you choose 3. Possiblities are:

a) Choosing the 3 other than green or

b) Choosing the 3 other than blue or

c) Choosing the 3 other than red or

d) Choosing the three other than adam.

Thus it is  4 posible ways. Or 4C3 = 4.

If you are interested in  permutation or arranging the  3 things chosen combination, the order also counts. Then for each of the 4 choices further arranging 3! whitin each choice makes the number of permutations 4P3 = 4*3! =24.

oceanberm | High School Teacher | (Level 1) eNoter

Posted on

Please seek out your question or teacher to find out whether order matters.  Certainly there is a math formula for this problem.  See below: If order does not matter :  answer = 24 Permutations are how many ways a given number of things can be sequenced. The standard notation is that nPn is the permutations (sequences) of all n things in a set of n things; nPr is the permutations on n things taken r at a time. The relevant formulas are below:   nPn = n! nPr = n!/(n - r)!     ABC, ACB, ADB, ABD, ADC, ACDBAC, BCA, BDA, BAD, BCD, BDCCAB, CBA, CBD, CDB, CAD, CDADAC, DCA, DAB, DBA, DBC, DCB   where ! is the factorial operation (n! = 1*2*3*...*n). For example, the number of permutations of 4 things taken 3 at a time is 4!/(4-3)! = 24/1 = 24. Thus for items, A, B, C, and D, the 24 permutations are: If order does matter : Answer = 4 Combinations refer to how many subsets can be derived from a given set, regardless of permutation sequence. For instance, ABC, ACB, BCA, BAC, CAB, and CBA are 6 permutations but only 1 combination. The relevant counting formulas are below:   nCn = 1 nCr = n!/(n - r)!r! By cancellation, n things combined 2 at a time = nC2 = n!/(n - 2)!2! = (n*(n-1))/2     A, B, and C in any orderA, B, and D in any orderA, C, and D in any orderB, C, and D in any order       For example, the number of combinations of 4 things taken 3 at a time is 4!/[(4-3)!3!] = 24/(1*6) = 4. The 4 combinations of A, B, C, and D are: Permutations and Combinations of Sets Not All Different. The foregoing formulas assumed that we were dealing with a set of n things, each of which was different from the other. But what if we have a set of, say, individuals, some of whom are Republicans and some Democrats, and we are interested in permutations by party, not individual? The relevant formulas are below:   nCn1, n2, ..., nk = 1 nPn1, n2, ..., nk = n!/n1!n2!...nk!