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

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

Answer :

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

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

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) = \(rac{2}{52}\) = \(rac{1}{26}\).

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

Description : 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

Last 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!) ... ways Hence, the required number of ways = 1! 51! 52! 3! (17!) 3 51!

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

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]

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

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}\) = \(rac{26 imes25 imes24}{52 imes51 imes50}\) = \(rac{2}{17}.\)

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

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 ... \(rac{4 imes3}{52 imes51}\) - \(rac{2}{52 imes51}\) = \(rac{660}{2652}\) = \(rac{55}{221}.\)

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

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 ... ) + \(rac{6}{1326}\) - \(rac{1}{1326}\) = \(rac{330}{1326}\) = \(rac{55}{221}\).

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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}\)= \(rac{44}{52}\) + \(rac{16}{52}\) = \(rac{28}{52}\) = \(rac{7}{13}\) .

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

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).

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

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.]

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

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

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

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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?

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

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

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

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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

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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

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Description : Among 15 players, 8 are batsman and 7 are bowlers. Find the probability that a team is chosen of 6 batsman and 5 bowlers ? -Maths 9th

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If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

Description : If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

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If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

Description : If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

Last Answer : Let AB and AC be two chords and AD be the diameter of the circle . Draw OL ⊥ AB and OM ⊥ AC . In ΔOLA and ΔOMA ∠ OLA = ∠ OMA = 90° OA = OA [common sides] ∠ OAL = ∠ OAM [given] . OLA ≅ OMA [by ASA congruency] OL =OM (by CPCT) Chords AM and AC are equidistant from O. ∴ AB = AC Hence proved.

A coin is tossed thrice and all eight outcomes are assumed equally likely. Find whether the events E -Maths 9th

Description : A coin is tossed thrice and all eight outcomes are assumed equally likely. Find whether the events E -Maths 9th

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

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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

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

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

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

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

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.

A bag contains a white and b black balls. Two players A and B alternately draw a ball from the bag -Maths 9th

Description : A bag contains a white and b black balls. Two players A and B alternately draw a ball from the bag -Maths 9th

Last Answer : (c) 2 : 1Let W denote the event of drawing a white ball at any draw and B that of drawing a black ball. Then, P (W) = \(rac{a}{a+b},\) P(B) = \(rac{b}{a+b}\)∴ P (A wins the game) = P (W or BBW or BBBBW or ... ... the given condition,\(rac{a+b}{a+2b}\) = 3. \(rac{b}{a+2b}\) ⇒ a = 2b ⇒ a : b = 2 : 1.

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

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:

Two people abi and krish invested in a business with 7 lakh and 8 lakh rupees respectively. They agree that 46% of the profit should be divided equally among them and rest is divided between them according to their investment. If krish got Rs. 2100 more than abi, Then the total profit is, A) 55000 B) 56000. C) 58333 D) None

Description : Two people abi and krish invested in a business with 7 lakh and 8 lakh rupees respectively. They agree that 46% of the profit should be divided equally among them and rest is divided between them according to their ... more than abi, Then the total profit is, A) 55000 B) 56000. C) 58333 D) None

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

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!

A game of the football with 11 players lasts for exactly 80minutes.There are 4 substitutes that alternate equally.If each player plays for the same length of the time,What is the duration? 1)120m 2)20m 3)84m 4)40m

Description : A game of the football with 11 players lasts for exactly 80minutes.There are 4 substitutes that alternate equally.If each player plays for the same length of the time,What is the duration? 1)120m 2)20m 3)84m 4)40m

Last Answer : 2)20m Exp: 80/4=20m.

The total agricultural land in a village is 1200 hectares. This is distributed among 320 families who form four groups in the following pattern. It is assumed that the land is distributed equally within each group. Identify the group of small farmers. Gr ou p Num ber of Tot al am ou nt of land Fam i l i es ow ned and oper at ed by each gr oup (i n hect ar es) A 100 300 B 180 300 C 30 300 D 10 300 (1) A (2) B (3) C (4) D

Description : The total agricultural land in a village is 1200 hectares. This is distributed among 320 families who form four groups in the following pattern. It is assumed that the land is distributed equally within each group. Identify the group of small ... B 180 300 C 30 300 D 10 300 (1) A (2) B (3) C (4) D

Last Answer : B 180 300

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

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.

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

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}\) = \(rac{25 imes24 imes23}{26 imes25 imes24}\) = \(rac{23}{26}\).

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

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.

What is the probability of drawing a jack and a queen consecutively from a deck of 52 cards, without replacement? a) 4/664 b) 8/52 c) 4/663 d) 4/52

Description : What is the probability of drawing a jack and a queen consecutively from a deck of 52 cards, without replacement? a) 4/664 b) 8/52 c) 4/663 d) 4/52

Last Answer : Answer: C)  Probability of drawing a jack = 4/52 = 1/13 After drawing one card, the number of cards are 51. Probability of drawing a queen = 4/51. Now, the probability of drawing a jack and queen consecutively is 1/13 * 4/51 = 4/663

Consider the example of finding the probability of selecting a red card or a 9 from a deck of 52 cards. A) 15/26 B) 26/15 C) 7/13 D) 13/7

Description : Consider the example of finding the probability of selecting a red card or a 9 from a deck of 52 cards. A) 15/26 B) 26/15 C) 7/13 D) 13/7

Last Answer : Answer: C) We need to find out P(R or 6) Probability of selecting a Red card = 26/52 Probability of selecting a 9 = 4/52 Probability of selecting both a red card and a 9 = 2/52  P(R or 9) = P(R) + P(9) – P(R and 9) = 26/52 + 4/52 – 2/52 = 28/52 = 7/13.

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

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

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?

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?

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There are 5 boys and 3 girls. In how many ways can they stand in a row so that no two girls are together? -Maths 9th

Description : There are 5 boys and 3 girls. In how many ways can they stand in a row so that no two girls are together? -Maths 9th

Last Answer : Have the 55 boys stand in a line. This can be done in 5!5! ways. For the moment, add a boy at each end, who will be removed when we're done. Now send the 33 girls, one at a time, to ... that each girl can either wedge herself between two boys or else stand next to a boy at one end or the other.

In how many ways can the letters of the word “AFLATOON” be arranged if the consonants and vowels must occupy alternate places? -Maths 9th

Description : In how many ways can the letters of the word “AFLATOON” be arranged if the consonants and vowels must occupy alternate places? -Maths 9th

Last Answer : 24 ways is the answer

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Description : In how many ways can 8 people sit around a circular Table? (a) 5040 (b) 40320 (c) 20160 (d) 2520 -Maths 9th

Last Answer : As is known in case of circular permutations, we keep one place fixed, so 8 people can sit around a circular table in (8 – 1) ! ways = 7! ways = 5040 ways.

A committee of 11 members sit at a round table. In how many ways can they be seated if the “President” and the “Secretary” choose to sit together. -Maths 9th

Description : A committee of 11 members sit at a round table. In how many ways can they be seated if the “President” and the “Secretary” choose to sit together. -Maths 9th

Last Answer : ∴ Required number of ways of seating = 9!

In how many ways can 7 men and 7 women sit on a round table such that no two women sit together? -Maths 9th

Description : In how many ways can 7 men and 7 women sit on a round table such that no two women sit together? -Maths 9th

Last Answer : Now there are 7 places vacant between these 7 men. ∴ 7 women can seat themselves in these 7 places in 7 ! ways. ∴ Total number of required arrangement where no two women sit together = 6!

In how many ways can 6 gentlemen and 3 ladies be seated round a table so that every gentleman may have a lady by his side? -Maths 9th

Description : In how many ways can 6 gentlemen and 3 ladies be seated round a table so that every gentleman may have a lady by his side? -Maths 9th

Last Answer : If each gentleman has to have a lady by his side, the seating arrangement can be done as shown below: This can be done in 5! (Gentlemen) × 3! (Ladies) = 720 ways.

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

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!

In how many ways can a committee of five persons be formed out of 8 members when a particular member is taken every time? -Maths 9th

Description : In how many ways can a committee of five persons be formed out of 8 members when a particular member is taken every time? -Maths 9th

Last Answer : ∴Required number of ways = 7C4 = 7!

A man has 6 friends. Number of different ways, he can invite 2 or more for dinner is -Maths 9th

Description : A man has 6 friends. Number of different ways, he can invite 2 or more for dinner is -Maths 9th

Last Answer : He can invite one or more friends by inviting 1 friend, or 2 friends or 3 friends, or all the 6 friends.

Out of 5 men and 2 women, a committee of 3 is to be formed. In how many ways can it be formed if at least one woman is to be included? -Maths 9th

Description : Out of 5 men and 2 women, a committee of 3 is to be formed. In how many ways can it be formed if at least one woman is to be included? -Maths 9th

Last Answer : Number of selections = Number of ways of selecting 2 men out of 5 men x number of ways of selecting l woman out of 3 women. = 5C2×3C1=5×41×2×3=30

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

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

A committee of 5 persons is to be formed out of 6 gents and 4 ladies. In how many ways can this be done if at most two ladies are included? -Maths 9th

Description : A committee of 5 persons is to be formed out of 6 gents and 4 ladies. In how many ways can this be done if at most two ladies are included? -Maths 9th

Last Answer : hope its clear