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A magician has K different cards each containing a distinct integer. T [#permalink]

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17 Sep 2017, 07:38

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A magician has K different cards each containing a distinct integer. The magician asks a participant to come on stage and chose n cards, where \(n<=K\). What is the probability that the cards chosen by the participant are in descending order or ascending order.

1. The magician has a set of 15 "K" cards in his hand 2. The participant choses 8 cards from the magician's hands.

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A magician has K different cards each containing a distinct integer. The magician asks a participant to come on stage and chose n cards, where \(n<=K\). What is the probability that the cards chosen by the participant are in descending order or ascending order.

1. The magician has a set of 15 "K" cards in his hand 2. The participant choses 8 cards from the magician's hands.

hi.. whatever be the number we choose ONLY one will be in descending order and one in ascending order. so total possibilities in all cases will be 2

but here we are looking for PROBABILITY, which is 2/n! where n is the number to be choosen. so we require only n for answer.

lets see the statements:-

1. The magician has a set of 15 "K" cards in his hand nothing about n insuff

2. The participant chooses 8 cards from the magician's hands. so prob = 2/8! suff

Re: A magician has K different cards each containing a distinct integer. T [#permalink]

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21 Sep 2017, 18:23

chetan2u wrote:

bkpolymers1617 wrote:

A magician has K different cards each containing a distinct integer. The magician asks a participant to come on stage and chose n cards, where \(n<=K\). What is the probability that the cards chosen by the participant are in descending order or ascending order.

1. The magician has a set of 15 "K" cards in his hand 2. The participant choses 8 cards from the magician's hands.

hi.. whatever be the number we choose ONLY one will be in descending order and one in ascending order. so total possibilities in all cases will be 2

but here we are looking for PROBABILITY, which is 2/n! where n is the number to be choosen. so we require only n for answer.

lets see the statements:-

1. The magician has a set of 15 "K" cards in his hand nothing about n insuff

2. The participant chooses 8 cards from the magician's hands. so prob = 2/8! suff

B

Hi Chentan2u

Thanks for the explanation, however I am still having trouble understanding how there will be 2 choices in all scenarios.

For instance, if the number of cards is X with X different integers (1...X). and we are to pick Y cards I believe there will be 3 different cases,

One in which we pick the lowest possible number (1) First, and then can ONLY pick ascending cards.

One in which we pick the highest possible number (X) first, and then can ONLY pick descending cards,

And finally, one in which we pick a card between 1 and X in which case we may have two viable choices either ascending or descending, which will only allow these two choices until one of the limits is hit.

For example, if we are to pick 1 through 10 as the number of cards, K and we are selecting 8 cards in either ascending or descending order, the probability of drawing a card anywhere in the middle is 8/10, however the probability of drawing successive ascending or descending cards changes whenever we hit the limit. If the first card we pick is 8, we have two choices, either 9 or 7, however if we pick 9, and 10 in the next two draws, we are left with ONLY 1 choice, which changes the probability of the draw.

I am sure my logic is flawed somewhere if you could help point or math guru Bunuel me in the right direction, I Would appreciate it.

A magician has K different cards each containing a distinct integer. The magician asks a participant to come on stage and chose n cards, where \(n<=K\). What is the probability that the cards chosen by the participant are in descending order or ascending order.

1. The magician has a set of 15 "K" cards in his hand 2. The participant choses 8 cards from the magician's hands.

hi.. whatever be the number we choose ONLY one will be in descending order and one in ascending order. so total possibilities in all cases will be 2

but here we are looking for PROBABILITY, which is 2/n! where n is the number to be choosen. so we require only n for answer.

lets see the statements:-

1. The magician has a set of 15 "K" cards in his hand nothing about n insuff

2. The participant chooses 8 cards from the magician's hands. so prob = 2/8! suff

B

Hi Chentan2u

Thanks for the explanation, however I am still having trouble understanding how there will be 2 choices in all scenarios.

For instance, if the number of cards is X with X different integers (1...X). and we are to pick Y cards I believe there will be 3 different cases,

One in which we pick the lowest possible number (1) First, and then can ONLY pick ascending cards.

One in which we pick the highest possible number (X) first, and then can ONLY pick descending cards,

And finally, one in which we pick a card between 1 and X in which case we may have two viable choices either ascending or descending, which will only allow these two choices until one of the limits is hit.

For example, if we are to pick 1 through 10 as the number of cards, K and we are selecting 8 cards in either ascending or descending order, the probability of drawing a card anywhere in the middle is 8/10, however the probability of drawing successive ascending or descending cards changes whenever we hit the limit. If the first card we pick is 8, we have two choices, either 9 or 7, however if we pick 9, and 10 in the next two draws, we are left with ONLY 1 choice, which changes the probability of the draw.

I am sure my logic is flawed somewhere if you could help point or math guru Bunuel me in the right direction, I Would appreciate it.

Hi..

Let's take total cards as 10 and you pick 4 cards out of them. These 4 could be anything, it does not matter. It could be 1,7,4,9 or 4,8,2,9 or 10,8,4,2 Pick any of these and you can arrange 1,7,4,9 in just one way of ascending 1,4,7,9 and one as descending 9,7,4,1 But how can these 4 be picked up - 4! Your answer 2/4!

It will be the case everywhere irrespective of the total 10 or which 4 you pick up
_________________

Re: A magician has K different cards each containing a distinct integer. T [#permalink]

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21 Sep 2017, 19:55

chetan2u wrote:

Hi..

Let's take total cards as 10 and you pick 4 cards out of them. These 4 could be anything, it does not matter. It could be 1,7,4,9 or 4,8,2,9 or 10,8,4,2 Pick any of these and you can arrange 1,7,4,9 in just one way of ascending 1,4,7,9 and one as descending 9,7,4,1 But how can these 4 be picked up - 4! Your answer 2/4!

It will be the case everywhere irrespective of the total 10 or which 4 you pick up

Thank you for the response Chentan2u, you cleared all my confusion.

I misunderstood and was thinking consecutive integers... If consecutive integers was the case would my explanation be valid?