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Q: What is the value of 4C2?
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How do you simplify 4c2 plus 5c plus 2c?

4c2 + 5c + 2c = 4c2 + 7c = c(4c + 7)


Factorization of b4 -16b2c2 - 80 c4?

(b2 - 20c2)(b2 + 4c2)


What is the product to 4C 2nd power times -2CD 2nd power?

4C2 * (-2CD2) = -8C3D2


What is 24c3 plus 375z3 factored completely?

24c3 + 375z3 = 3*(8c3 + 125z3) = 3*(2c + 5z)*(4c2 - 10cz + 25z2)


What is the probability of tossing 4 pennies and landing on heads twice and tails twice?

4C2(1/2)4 = 6/16 = 3/8


How do you solve 4 cosine squared x - 2 equals 0?

Let c represent cos x. Then, we have, 4c2 - 2 = 0; whence, 4c2 = 2, c2 = ½, and c = ±√½ = ±½√2; that is, cos x = ±½√2. From this, it is evident that, for 0 ≤ x < 360°, x = 45°, 135°, 225°, or 315°; or, if you prefer, for 0 ≤ x < 2π, x = ¼π, ¾π, 1¼π, or 1¾ π.


How many permutations of two different digits can be obtained from a set of four different digits?

6 of them. 4C2 = 4!/(2!*2!) = 4*3/(2*1) = 6


What is the probability of getting 2 aces when given 5 cards out of a normal 52 card deck?

5 cards out of 52 may be drawn in 52C5 = 2 598 960 ways. 2 aces out of 4 may be drawn in 4C2 = 6 ways. 3 non-aces out of 48 non-aces may be drawn in 48C3 = 17 296 ways P( 2 aces ) = (4C2)(48C3) / (52C5) = (6)(17 296)/( 2 598 960) = 0.0399


How many parallelograms are formed when a set of 5 parallel lines intersect a set of 4 parallel lines?

This one is much more straightforward. There are 5C2 = 10 ways to choose two parallel lines from the set of five. There are 4C2 = 6 ways to choose two parallelograms from a set of four. Any parallelogram is uniquely determined by one pair of lines from the five, and one pair of lines from the four. Thus, the number of possible parallelograms is(5C2)*(4C2) = (10)*(6) = 60


How do you factor this binomial 24cĀ² plus 96c plus 90?

24c2 + 96c + 90 = 6*(4c2 + 16c + 15) = 6*(2c + 3)*(2c + 5)


In how many ways can a 3 person committee be selected from four distinct pairs of husband and wives so as NOT to have spouse in it?

let the pairs be named as A1 B1,A2 B2,A3 B3,A4 B4.LET US SELECT TWO FROM BOYS AND ONE FROM GIRL THIS WE CAN DO IN 4c2*2c1=12 NOW THE OTHER WAY i.e 2 GIRLS AND 1 BOY 4c2*2c1=12 OTHER WAY IS TO SELECT ALL THE THREE FROM GIRLS OR BOYS=2*4c3 TOTAL 30


A bag contain 4 balls 2 balls are drawn at random and are found to be white What is the probability that all balls are white?

it has three cases 1st E1- two balls are whiteE2- three balls are whiteE4-four balls are whiteE4/w= (p(E3)*4C2/4C2)/(P(E1)*2C2/4C2+P(E2)3C2/4C2+P(E3)*4C2/4C2)WHERE-P(E1)=P(E2)=P(E3)=1/3----------------------------------------------------------------------------------------------------2nd opinionLet's say we have 3 boxes with 4 balls each.Box A has 4 white balls.Box B has 3 white balls.Box C has 2 white balls.The probability of drawing 2 W balls from;Box A, P(2W│A)=(4/4)∙(3/3)=1Box B, P(2W│B)=(3/4)∙(2/3)=1/2Box C, P(2W│C)=(2/4)∙(1/3)=1/6Say the probability of picking any of the 3 boxes is the same, we have;P(A)=1/3P(B)=1/3P(C)=1/3Question is, given the event of drawing 2 W balls from a box taken blindlyfrom the 3 choices, what is the probability that the balls came from box A,P(A│2W).Recurring to Bayes Theorem:P(A│2W)=[P(A)P(2W│A)]/[P(A)P(2W│A)+P(B)P(2W│B)+P(C)P(2W│C)]=[(1/3)(1)]/[(1/3)(1)+(1/3)(1/2)+(1/3)(1/6)]=6/10=0.60=60%P(A│2W)=0.60=60%Read more:Solution_to_a_bag_contains_4_balls_Two_balls_are_drawn_at_random_and_are_found_to_be_white_What_is_the_probability_that_all_balls_are_white