In how many ways can a pack of 52 cards be divided into 4 sets, three of them having 16 cards each and the fourth just 4 cards? -Maths 9th

1 Answer

Answer :

First we divide 52 cards into two sets which contains 1 and 51 cards respectively is 1! 51! 52! ​ Now 51 cards can be divided equally in three sets each contains 17 cards (Here order of sets is not important) in 3!(17!) 3 52! ​ ways Hence, the required number of ways = 1! 51! 52! ​ × 3! (17!) 3 51! ​

Related questions

Description : In how many ways can a pack of 52 cards be divided equally among four players in order? -Maths 9th

Last Answer : Distribution of 52 cards can be equally divided among four players. Hence, number of ways is (13!)4! 52! ​ 4!= (13!) 52! ​

Description : One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. -Maths 9th

Last Answer : (c) \(rac{1}{26}\)There is a total of 52 cards n(S) = 52 Let A : Event of drawing a red king Since there are only two red kings in the pack, n(A) = 2 ∴ P(A) = \(rac{2}{52}\) = \(rac{1}{26}\).

Description : Find the probability that the three cards drawn from a pack of 52 cards are all black ? -Maths 9th

Last Answer : Number of ways in which three cards can be drawn from a pack of 52 cards n(S) = 52C3. Let A : Event of drawing all the three cards as black Then, n(A) = 26C3 (∵There are 26 black cards)∴ P(A ... (rac{^{26}C_3}{^{52}C_3}\) = \(rac{26 imes25 imes24}{52 imes51 imes50}\) = \(rac{2}{17}.\)

Description : Two cards are drawn from a well shuffled pack of 52 cards one after another without replacement. -Maths 9th

Last Answer : Probability of drawing an ace in the first draw = \(rac{4}{52}.\)Probability of drawing a queen of opposite shade in the second draw = \(rac{2}{51}.\)Probability of drawing a queen in the first draw = \(rac{4}{52}.\) ... \(rac{2}{51}\) = \(rac{4}{663}.\) [ AND' and OR'Theorems]

Description : Two cards are drawn from a pack of 52 cards. What is the probability that either both are red or both are kings ? -Maths 9th

Last Answer : Let S : Drawing 2 cards out of 52 card A : Drawing 2 red cards B : Drawing 2 kings A ∪ B : Drawing 2 red cards or 2 kings ∴ n(S) = 52C2 n(A) = 26C2 (∵ There are 26 red cards) n(B) = 4C2 ... \(rac{4 imes3}{52 imes51}\) - \(rac{2}{52 imes51}\) = \(rac{660}{2652}\) = \(rac{55}{221}.\)

Description : Two cards are drawn at random from a well-shuffled pack of 52 cards. What is the probability that either both are black or both are kings ? -Maths 9th

Last Answer : (b) \(rac{55}{221}\)S : Drawing 2 cards out of 52 cards ⇒ n(S) = 52C2 = \(rac{|\underline{52}}{|\underline{52}|\underline2}\) = \(rac{52 imes51}{2}\) = 1326A : Event of drawing 2 black cards out of 26 black cards⇒ n ... ) + \(rac{6}{1326}\) - \(rac{1}{1326}\) = \(rac{330}{1326}\) = \(rac{55}{221}\).

Description : A card is drawn at random from a well-shuffled pack of 52 cards. Find the probability of getting a red card or a diamond or a jack ? -Maths 9th

Last Answer : (d) \(rac{7}{13}\)Here n(S) = 52 Let A, B, C be the events of getting a red card, a diamond and a jack respectively. ∵ There are 26 red cards, 13 diamonds and 4 jacks, n(A) = 26, n(B) = 13, n(C) = 4 ⇒ n(A ∩ B) = ... rac{1}{52}\)= \(rac{44}{52}\) + \(rac{16}{52}\) = \(rac{28}{52}\) = \(rac{7}{13}\) .

Description : A card is drawn at random from a well shuffled pack of 52 cards -Maths 9th

Last Answer : (c) P(X) = P(Y) > P(Z) P(X) = \(rac{26}{52}\) + \(rac{4}{52}\) - \(rac{2}{52}\) = \(rac{28}{52}\) (∵ There are 26 black cards, 4 kings and 2 black kings)P(Y) = \(rac{13}{52}\) + \(rac{ ... }{52}\)(∵ There are 4 aces, 13 diamonds, 4 queens, 1 ace of diamond, 1 queen of diamond) ∴ P(X) = P(Y) > P(Z).

Description : Two players A and B play a game by alternately drawing a card from a well-shuffled pack of playing cards, replacing the card each time after draw. -Maths 9th

Last Answer : (a) \(rac{13}{25}\)Let E : Event of drawing a queen in a single draw the pack of 52 cards. As there are 4 queens in a pack of 52 cards,P(E) = \(rac{4}{52}\) = \(rac{1}{13}\)P(\(\bar{E}\)) = P(not ... {25}\). [Sum of a G.P with infinite terms = \(rac{a}{1-r}\) where a = 1st term, r = common ratio.]

Description : Two cards are drawn at random from a pack of 52 cards. What is the probability that both of them are either black or queen cards? a) 55/442 b) 54/221 c) 55/221 d) 51/221

Last Answer : c) 55/221

Description : Four cards are drawn from a full pack of cards. Find the probability that : -Maths 9th

Last Answer : 4 cards can be drawn from a pack of cards in 52C4 ways ∴ Exhaustive number of cases = n(S) = 52C4 (a) There are 4 suits, each containing 13 cards. Let A : Event of drawing one card from each suit ⇒ Favourable number of ... = \(rac{15229}{54145}\) (∵ P(Event) + P(complement of event) = 1)

Description : A pack of cards contains 4 aces, 4 kings, 4 queens and 4 jacks. -Maths 9th

Last Answer : (d) \(rac{9}{20}\)Let S be the sample space for drawing 2 cards out of 4 aces, 4 kings, 4 queens and 4 jacks i.e, 16 cards. Then n(S) = 16C2 P(Drawing at least one ace) = 1 - P(Drawing no ace) Let E : Event of ... \(rac{11}{20}\)∴ P(drawing at least one ace) = 1 - \(rac{11}{20}\) = \(rac{9}{20}\) .

Description : How many ways can 52 deck cards be divided into equal piles?

Last Answer : 3 ways

Description : How many ways can 52 deck cards be divided into equal piles?

Last Answer : 3 ways

Description : What is the probability a well shuffled pack of 52 cards a card is drawn at random find the probability that it is either a heart or a queen?

Last Answer : 41365

Description :  Two cards are drawn at random from a pack of 52 cards.what is the probability that either both are Red or both are king? A) 52/221 B) 55/190 C) 55/221 D) 19/221

Last Answer : Answer: C) We have n(s) = 52C2 = 1326. Let A = event of getting both red cards B = event of getting both king A∩B = event of getting king of red cards n(A) = 26C2 = 325, n(B)= 4C2= 6 and n(A∩B) = 2C2 = 1 P(A ... S) = 1/1326 P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

Description : The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then, the area of the triangle is -Maths 9th

Last Answer : The area of the triangle is

Description : In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Last Answer : NEED ANSWER

Description : The sides of a triangle are 56 cm, 60 cm and 52 cm long. Then, the area of the triangle is -Maths 9th

Last Answer : The area of the triangle is

Description : In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Last Answer : Median will be a good representative of the data, because 1.each value occurs once. 2.the data is influenced by extreme values.

Description : Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five. In how many different ways -Maths 9th

Last Answer : According to the question, we have 5 balls to be placed in 3 boxes where no box remains emptyHence, we can have the following kinds of distribution firstly, where the distribution will be (3,1,1) that is, one box gets three ... go in second box is = 4 C 2 . Total no. of ways =90 Total:60+90=150.

Description : A cylindrical rod of iron whose radius is one-fourth of its height is melted and cast into spherical balls of the same radius as that of the cylinder. -Maths 9th

Last Answer : Let radius of cylindrical rod =r ⇒ height =4r Volume of cylindrical rod =πr2h =πr2(4r) =4πr3 Volume of spherical balls of radius r=34​πr3 No. of balls =34​πr34πr3​=

Description : Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm x 8 cm x 8 cm. -Maths 9th

Last Answer : Volume of rectangular box=lbh=16(64)=1024cm3 Volume of sphere=34​πr3=33.5238cm3 16 sphere=16(33.5238)=536.3808 Volume of liquid=1024−536.3808=488cm3

Description : Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm x 8 cm x 8 cm. -Maths 9th

Last Answer : According to question find the volume of this liquid.

Description : A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Last Answer : since curved surface of half of the spherical ball = 56.57 cm2 ∴ 2πr2 = 56.57 ⇒ r2 = 56.57 / 2 × 3.14 = 9 ⇒ r = 3 cm Now, volume of spherical ball = 4 / 3 πr3 = 4 / 3 × 3.14 × 3 × 3 × 3 = 113.04 cm3

Description : A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Last Answer : since curved surface of half of the spherical ball = 56.57 cm2 ∴ 2πr2 = 56.57 ⇒ r2 = 56.57 / 2 × 3.14 = 9 ⇒ r = 3 cm Now, volume of spherical ball = 4 / 3 πr3 = 4 / 3 × 3.14 × 3 × 3 × 3 = 113.04 cm3

Description : If a circle is divided into eight equal parts, find the angle subtended by each arc at the centre. -Maths 9th

Last Answer : Total angle at centre = 360° When divided into eight part, Angle subtended by each arc = 360 / n8 = 45°

Description : The diameter of the moon is approximately one fourth of the diameter of the earth. Find the ratio of their surface areas -Maths 9th

Last Answer : If diameter of earth is said d, then the diameter of moon will be d/4 (as per given statement) Radius of earth = d/2 Radius of moon = ½×d/4 = d/8 Surface area of moon = 4π(d/8)2 Surface area of earth = 4π(d/2)2 Ncert solutions class 9 chapter 13-6

Description : If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Last Answer : (b) In point P (-1, 1), x-coordinate is -1 unit and y-coordinate is 1 unit, so it lies in llnd quadrant. Similarly, we can plot all the points Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4), It is clear from the graph that points R and Q lie in fourth quadrant.

Description : The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Last Answer : Solution of this question

Description : If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Last Answer : (b) In point P (-1, 1), x-coordinate is -1 unit and y-coordinate is 1 unit, so it lies in llnd quadrant. Similarly, we can plot all the points Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4), It is clear from the graph that points R and Q lie in fourth quadrant.

Description : The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to -Maths 9th

Last Answer : Solution of this question

Description : A fair dice is thrown twenty times. The probability that on the tenth throw the fourth six appears is -Maths 9th

Last Answer : (c) \(rac{84 imes5^6}{6^{10}}\)In the first nine throws we should have three sixes and six non-sixes and a six in the tenth throw and thereafter whatever face appears, it doesn't matter. ∴ Required probability = 9C3 \(\bigg(rac{1} ... x 1 x 1 ............x 1 {10 times} = \(rac{84 imes5^6}{6^{10}}\).

Description : If A(3, 5), B(– 5, – 4), C(7, 10) are the vertices of a parallelogram taken in order, then the co-ordinates of the fourth vertex are: -Maths 9th

Last Answer : (c) RhombusCo-ordinates of P are \(\bigg(rac{-1-1}{2},rac{-1+4}{2}\bigg)\)i.e, \(\big(-1,rac{3}{2}\big)\)Co-ordinates of Q are \(\bigg(rac{-1+5}{2},rac{4+4}{2}\bigg)\)i.e, (2, 4)Co-ordinates of R ... \sqrt{(2-2)^2+(4+1)^2}\) = \(\sqrt{25}\) = 5⇒ PR ≠ SQ ⇒ Diagonals are not equal ⇒ PQRS is a rhombus.

Description : A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided ? -Maths 9th

Last Answer : From the adjoining figure, we have The field PQRS is divided into three parts △PAQ, △APS and △AQR. Now, △PAQ and ||gm PQRS are on the same base and lie between the same parallels. ∴ ar(△PAQ) = 1 / 2 ar(||gm ... , she can sow wheat in △APS and △AQR, pulses in △PAQ or vice - versa .

Description : A farmer was having a field in the form of a parallelogram PQRS. She took any point A on RS and joined it to points P and Q. In how many parts the field is divided ? -Maths 9th

Last Answer : From the adjoining figure, we have The field PQRS is divided into three parts △PAQ, △APS and △AQR. Now, △PAQ and ||gm PQRS are on the same base and lie between the same parallels. ∴ ar(△PAQ) = 1 / 2 ar(||gm ... , she can sow wheat in △APS and △AQR, pulses in △PAQ or vice - versa .

Description : All red face cards are removed from a pack of playing cards. The remaining cards are well shuffled and then a card is drawn at random from them. Find the probability that the drawn card is face card. a. 2/23 b. 7/44 c. 3/23 d. 4/25

Last Answer : c. 3/23

Description : Three cards are drawn at random from an ordinary pack of cards. Find out the probability that they will consist of a king, aqueen and an ace?

Last Answer : Answer: 64/2210.

Description : Cards with numbers 1, 2, 3, ........... 100 are -Maths 9th

Last Answer : (i) Favourable cards are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, i.e., 10. P (prime number less than 30) = 10/100 or 1/10 (ii) Favourable cards are 35, 70. P (card is a multiple of 5 and 7) = 2/100 or 1/50 ( ... 42, 49, 56, 63, 77, 84, 91, 98 i.e., 32 P(card is a multiple of 5 or 7) = 32/100 or 8/25.

Description : Two decks of playing cards are well shuffled and 26 cards are randomly distributed to a player. -Maths 9th

Last Answer : (b) \(rac{23}{26}\)Total number of ways in which 3 letters can be selected from 26 letters = 26C3. If A is not to be included in the choice, there are 25 letters left, so number of ways in which 3 letters can be ... 25}C_3}{^{26}C_3}\) = \(rac{25 imes24 imes23}{26 imes25 imes24}\) = \(rac{23}{26}\).

Description : If Pepsi sets the price of its six pack to match exactly the price of Coca - Cola's, Pepsi is using a ------ pricing method. A)demand -oriented B)cost -oriented C)experience curve D)competition -oriented

Last Answer : D)competition -oriented

Description : Four friends have 7 shirts, 6 pants and 8 ties. In how many ways can they wear them? -Maths 9th

Last Answer : 7 shirts can be worn by 4 friends in 7P4 ways. Similarly, 6 pants and 8 ties can be worn by 4 friends in 6P4 and 8P4 ways respectively. ∴ Total number of ways in which 7 shirts, 6 pants and 8 ties can be worn by 4 friends = 7P4 × 6P4 × 8P4 = 7! 3!

Description : In how many ways can a team of 11 players be selected from 14 players when two of them play as goalkeepers only? -Maths 9th

Last Answer : As each team of 11 players has one goalkeeper and 10 team members, and out of 14 players there are 2 goalkeepers and 12 team members. = 12×112×2 = 132.

Description : 7 relatives of a man comprise 4 ladies and 3 gentlemen; his wife also has 7 relatives; 3 of them are ladies and 4 gentlemen. In how many ways -Maths 9th

Last Answer : The possible cases are: Case I : A man invites (3 ladies) and woman invites (3 gentlemen) 4 C 3 4 C 3 =16 Case II : A man invites (2 ladies,1 gentleman) and woman invites (2 gentleman, 1 lady) ( 4 ... and woman invites (3 ladies) 3 C 3 3 C 3 =1 Total number of ways =16+324+144+1=485

Description : According to I.S. : 456, slabs which span in two directions with corners held down, are assumed to be divided in each direction into middle strips and edge strips such that the width of the middle strip, is ... (C) Three-fourth of the width of the slab (D) Four-fifth of the width of the slab

Last Answer : Answer: Option C

Description : Find the number of ways in which 10 different flowers can be strung to form a garland so that three particular flowers are always together -Maths 9th

Last Answer : Consider the three particular flowers as one flower. Then we have (10 – 3) + 1 = 8 flowers which can be strung in the garland. Thus the garland can be formed in (8 – 1)!, i.e., 7! ways But the 3 particular flowers can be arranged amongst themselves in 3!

Description : Three identical dice are rolled. The probability that same number will appear on each of them is -Maths 9th

Last Answer : (d) \(rac{1}{36}\)Total number of outcomes when three identical dice are rolled, n(S) = 6 6 6 = 216 Let A : Event of rolling same number or each dice ⇒ A = {(1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), ... 6)} ⇒ n(A) = 6 ∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{6}{216}\) = \(rac{1}{36}\).

Description : Three telephone circuits, each having an S/N ratio of 44 dB, are connected in tandem. What is the overall S/N ratio? A. 44 dB B. 39.2 dB C. 41 dB D. 43.52 dB

Last Answer : B. 39.2 dB

Description : In a chess tournament, where participants were to play one game with one another, two players fell ill, having played 6 games each without playing -Maths 9th

Last Answer : answer:

Description : Do playing card manufacturers print all 52 (or 54) cards at once?

Last Answer : I do not know, but I would think that they print all cards onto a single sheet of cardboard, then cut them out. Otherwise, you would need a separate printer for each card, which seems ludicrous to me.