SOLUTION: i need your help with this question.... evalutate C(7, 4) and this question also... evaluate 5!

Question 259703: i need your help with this question.... evalutate C(7, 4) and this question also... evaluate 5!
Found 2 solutions by stanbon, Theo:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
evaluate C(7, 4) and this question also... evaluate 5!
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C(7.4) = 7!/[(7-4)!*4!] = (7*6*5*4]/[1*2*3*4] = 35
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5! = 5*4*3*2*1 = 120
==========================
Cheers,
Stan H.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
c(7,4) is the possible combinations of 4 things out of 7.

you have 7 elements in the set.

you draw 4 of them.

how many possible combinations can you make?

combinations are not permutations, so order is not important.

this means that abcd is considered to be the same set as dcba since they contain the same elements only in a different order.

the formula for number of combinations is:

c(n,x) = n! / (x!*(n-x)!)

in your example, this becomes:

c(7,4) = 7! / (4!*3!) which becomes:

c(7,4) = (7*6*5*4*3*2*1) / (4*3*2*1*3*2*1)

the 4*3*2*1 in the numerator and denominator cancels out and you are left with:

c(7,4) = (7*6*5)/(3*2*1)

the 6 in the numberator and the (3*2) in the denominator cancel out and you are left with:

c(7,4) = (7*5) = 35.

the possible number of combinations of 4 element sets out of a 7 element set is equal to 35.

5! is equal to 5*4*3*2*1 = 20*3*2*1 = 60*2*1 = 120.

on the c(7,4) question, to see what is happening, try a smaller number of combinations from a smaller set.

try c(3,2)

this is the number of possible sets of 2 elements out of a set of 3 elements.

the formula would tell you:

c(3,2) = 3!/(2!1!) = (3*2*1)/(2*1*2) = 3

you can very easily see that this is true by taking abc as the 3 elements in the set.

draw all possible different sets of 2 out of this to get:

ab
ac
bc

since these are combinations, order is not important, so ba is the same set as ab, ca is the same set as ac, cb is the same set as bc.

if it were permutations rather than combinations, then the possible number of sets would be 6.

that would be:

ab
ba
ab
ca
bc
cb

in that case the formula would have been p(3,2) = n! / (n-x)!.

the x! would have been removed from the denominator of the formula.

p(3,2) = 3!/1! = 3*2*1 / 1 = 6.

combination formula is:

c(n,x) = n! / (x!*(n-x)!)

c(3,2) = 3! / (2!*1!)

permutation formula is:

p(n,x) = n! / (n-x)!

p(3,2) = 3! / 1!

you were looking for c(7,4), so your answer is 35 as stated above.

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