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4!=4*3*2*1 2!=2*1 4!/2!=4*3=12
= 1*2*3*4 / 1*2 1 & 2 - cancel out so: =3*4/1=12
the number of ways to pick two out of 4 is (4 2) =6
yeah that is 4!/(2!2!)
how its double 2.
do you know the equation for nCr?
c means combination.
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
do you want to try to compute 6 choose 2 and I'll tell you if it's right?
4!/2! = 4x3x2!/2! = 4x3 = 12
is this not same as combination..
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
combination of 4 choose 2 is 4!/(2!2!)=6
if we have to calculate to " how many ways can it be done" which method should be applied?
it depends on the question
do you have a specific problem
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.
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
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
does that make sense?
how do solve for (7 c 3)
am the student of medicine so dont have an idea on math.
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
(7*6*5/3*2*1) is it right?
but urs is different from it isnt it?
no it is the same. 7!=7*6*5*4! so the 4! goes away
(7*6*5/3*2*1) how it becames.....shouldnt it be like (7*6*5*4*3*2/3*2*1) ;)
no. 7*6*5*4!/(4!*3*2*1) 4! cancels. 7*6*5/(3*2*1)
3*2*1=6 so 6 cancels as well. 7*5=35= (7 c 3)
in combination we should always use the formula n c r= n!/(n-r)*r
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!
is it necessary to write r! as well after n-r! r! comes in n-r! isnt it?
anyway i got the things that i lost.thanks
the equation needs both. n!/((n-r)!*r!)
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?