Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

right?

Answer 5

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

Answer 6

how its double 2.

Answer 7

do you know the equation for nCr?

Answer 8

not exactly.

Answer 9

c means combination.

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

is this not same as combination..

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

it depends on the question

Answer 19

do you have a specific problem

Answer 20

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.

Answer 21

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

Answer 22

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

Answer 23

does that make sense?

Answer 24

how do solve for (7 c 3)

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

yep

Answer 29

but urs is different from it isnt it?

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

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

Answer 35

right?

Answer 36

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!

Answer 37

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

Answer 38

anyway i got the things that i lost.thanks

Answer 39

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

Answer 40

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?