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anonymous
 5 years ago
am confused in simple matter 4!/2! can anyone explain it plz?
anonymous
 5 years ago
am confused in simple matter 4!/2! can anyone explain it plz?

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.04!=4*3*2*1 2!=2*1 4!/2!=4*3=12

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0= 1*2*3*4 / 1*2 1 & 2  cancel out so: =3*4/1=12

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the number of ways to pick two out of 4 is (4 2) =6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah that is 4!/(2!2!)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you know the equation for nCr?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0when 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 (nr)!*(r!) which is the number being chosen and the number not being chosen

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nCr(n,r)=n!/((nr)!r!)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you want to try to compute 6 choose 2 and I'll tell you if it's right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.04!/2! = 4x3x2!/2! = 4x3 = 12

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is this not same as combination..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0combination 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
 5 years ago
Best ResponseYou've already chosen the best response.0combination of 4 choose 2 is 4!/(2!2!)=6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if we have to calculate to " how many ways can it be done" which method should be applied?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it depends on the question

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you have a specific problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0from 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
 5 years ago
Best ResponseYou've already chosen the best response.0so 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
 5 years ago
Best ResponseYou've already chosen the best response.0so 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
 5 years ago
Best ResponseYou've already chosen the best response.0does that make sense?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do solve for (7 c 3)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0am the student of medicine so dont have an idea on math.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm a graduate math student and I dont know anything about medicine haha. (7 c 3)= 7!/((73)!*3!)=7!/(4!3!)=35

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(7*6*5/3*2*1) is it right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but urs is different from it isnt it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no it is the same. 7!=7*6*5*4! so the 4! goes away

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(7*6*5/3*2*1) how it becames.....shouldnt it be like (7*6*5*4*3*2/3*2*1) ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no. 7*6*5*4!/(4!*3*2*1) 4! cancels. 7*6*5/(3*2*1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.03*2*1=6 so 6 cancels as well. 7*5=35= (7 c 3)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0in combination we should always use the formula n c r= n!/(nr)*r

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you are choosing r objects out of n and order does not matter . the two terms in the denominator are factorial as well. n!/(nr)!r!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is it necessary to write r! as well after nr! r! comes in nr! isnt it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0anyway i got the things that i lost.thanks

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the equation needs both. n!/((nr)!*r!)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0would 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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