Seven cards are marked with the letters A to G. They seven cards are

Question

Seven cards are marked with the letters A to G. They seven cards are shuffled randomly and then placed in a single stack. What is the probability that the cards marked A and G are not adjacent in the stack?

A. 1/84
B. 1/7
C. 2/7
D. 5/7
E. 6/7

A. 1/84
B. 1/7
C. 2/7
D. 5/7
E. 6/7

Aks111 · Answer

7 cards can be arranged in 7! ways.
P(A and G not adjacent) = 1 - P(A and G are adjacent)

If A and G are considered as 1 unit and shuffled with remaining 5 cards, 
Total # of ways of shuffling 5 cards and 1 unit = 6!
Since the unit i.e. A and G can also be shuffled among themselves in 2! ways,
Total # of ways of shuffling the cards such that A and G are adjacent = 6! * 2!
P(A and G are adjacent) = (6! * 2!) / 7! = 2/7

P(A and G not adjacent) = 1- 2/7 = 5/7
 Answer: D

Aks111 · Answer

An individual probability based calculation:
P(A and G are not adjacent) = P(A in 1st place and G not in 2nd place ) +
P(A in 2nd place and G not in 1st or 3rd) + P(A in 3rd place and G not in 2nd or 4th place) +...
P(A in 7th place and G not in 6th place)

P(A is in 1st place and G is not in 2nd place) = P(A in 1st place) * P(G not in 2nd place)
P(A) in 1st place = 1/7
P(G) not in the 2nd place i.e. G is in 3,4,5,6,7 i.e. G has 5 available places out of 6 = 5/6
P(A is in 1st place and G is not in 2nd place) = 1/7 * 5/6 ...........(I)
This scenario is same as P(A is in 7th place and G is not in 6th place) .........(II)

P(A in 2nd place) = 1/7
P(G) not in the 1st or 3rd place = P(G is in 4,5,6 or 7 place) = G has 4 available places out of 6 = 4/6 ------(III)
P(A in 2nd place and G not in 1st or 3rd) = 1/7 * 4/6
This scenario is same as A in 3rd, 4th, 5th or 6th place. .......(IV)

Therefore, P(A and G are not adjacent) = (I) + (II) + (III) + 4* (IV) = 2* 1/7 * 5/6 + 5* 1/7* 4/6
= 5/7

GMATWhizTeam · Answer

Solution:

We are given 7 cards marked A to G. These cards are kept together after shuffling.

We are asked  P(A and G not kept together)

For this, we will need the number of arrangements when A and G are not kept together. This is huge and may take a lot of time.

For questions like these, we should apply the methodology " Number of favorable cases = Total cases - Number of unfavorable cases ".

In our question, the Number of unfavorable cases = cases where A and G are kept together  = 6! 	imes 2! . 6! because we are taking A and G as a single entity and 2! because they can arrange themselves as well.

Now,  Number of favorable cases = Total cases - Number of unfavorable cases  = 7!-6!	imes 2!

Thus, P(A and G not kept together)  = \frac{7!-6!	imes 2!}{7!}=1-\frac{6!	imes 2!}{7!}=1-\frac{2}{7}=\frac{5}{7}

Hence the right answer is    Option D   .

BAA5 · Answer

Total number of arrangements of 7 cards = 7!
Arrangements with A and G together = 6!* 2 ( 2 arrangements within A and G - taken as a single unit)
Total number of arrangements where A and G are not together 7!-2*6! = 7*6! - 2*6! = 6!(7-2) = 5*6!

Probability = 5*6!/7! = 5/7

bumpbot · Answer

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Seven cards are marked with the letters A to G. They seven cards are

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