am confused in simple matter 4!/2! can anyone explain it plz?

- anonymous

am confused in simple matter 4!/2! can anyone explain it plz?

- katieb

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- anonymous

4!=4*3*2*1
2!=2*1
4!/2!=4*3=12

- anonymous

= 1*2*3*4 / 1*2
1 & 2 - cancel out
so:
=3*4/1=12

- anonymous

the number of ways to pick two out of 4 is
(4 2) =6

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## More answers

- anonymous

right?

- anonymous

yeah that is 4!/(2!2!)

- anonymous

how its double 2.

- anonymous

do you know the equation for nCr?

- anonymous

not exactly.

- anonymous

c means combination.

- anonymous

when you're choosing a certain number of objects from a total you partition into a group being chosen and a group not being chosen. so n choose r we have n! total possibilities and we have to divide by (n-r)!*(r!) which is the number being chosen and the number not being chosen

- anonymous

nCr(n,r)=n!/((n-r)!r!)

- anonymous

do you want to try to compute 6 choose 2 and I'll tell you if it's right?

- anonymous

4!/2! = 4x3x2!/2! = 4x3 = 12

- anonymous

is this not same as combination..

- anonymous

combination where order does not matter is nCr and I posted that formula above. 4!/2! is not in the standard form for a combination

- anonymous

combination of 4 choose 2 is 4!/(2!2!)=6

- anonymous

if we have to calculate to " how many ways can it be done" which method should be applied?

- anonymous

it depends on the question

- anonymous

do you have a specific problem

- anonymous

from a grp of 7 men and 6 women, 5 persons are to be selected to form a comitte so that at least 3 men are there on comittee. in how many way can it be done.

- anonymous

so there are three cases, you choose 3 4 or 5 men.
choose three men: (7 c 3)*(6 c 2)
choose 4: (7 c 4)*(6 c 1)
choose all 5 men: (7 c 5)
Then you would add all three cases together

- anonymous

so basically you're choosing your men out of 7 * the remaining people you need in the comittee chosen form the total number of girls

- anonymous

does that make sense?

- anonymous

how do solve for (7 c 3)

- anonymous

am the student of medicine so dont have an idea on math.

- anonymous

I'm a graduate math student and I dont know anything about medicine haha. (7 c 3)= 7!/((7-3)!*3!)=7!/(4!3!)=35

- anonymous

(7*6*5/3*2*1) is it right?

- anonymous

yep

- anonymous

but urs is different from it isnt it?

- anonymous

no it is the same. 7!=7*6*5*4! so the 4! goes away

- anonymous

(7*6*5/3*2*1) how it becames.....shouldnt it be like (7*6*5*4*3*2/3*2*1) ;)

- anonymous

no. 7*6*5*4!/(4!*3*2*1)
4! cancels. 7*6*5/(3*2*1)

- anonymous

3*2*1=6 so 6 cancels as well. 7*5=35= (7 c 3)

- anonymous

in combination we should always use the formula n c r= n!/(n-r)*r

- anonymous

right?

- anonymous

if you are choosing r objects out of n and order does not matter . the two terms in the denominator are factorial as well. n!/(n-r)!r!

- anonymous

is it necessary to write r! as well after n-r! r! comes in n-r! isnt it?

- anonymous

anyway i got the things that i lost.thanks

- anonymous

the equation needs both. n!/((n-r)!*r!)

- anonymous

would you plz solve this : a man has four socks in his drawer . each socket is either black or white . the changes of him selecting a pair at random and finding that he has a white pair is 0.5. what one his chances of the pair being black?

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