7 white balls and 3 black balls are placed in a row at random. The pro

The black ball will not be adjacent if there is a white ball in the middle.

First, place all the 7 white balls, and there will be 8 spaces available for 3 black balls.

Desired result: \(^{8}\mathrm{C_3}\)

Total result: \(^{10}\mathrm{C_3}\)

Probability: \(\frac{Desired}{ Total}\)

=> {8*7*6} / {10*9*8}

=> \(\frac{7}{15}\)

Answer B

Solution:

The number of ways 7 white balls and 3 black balls can be placed in a row is 10! / (7!3!) = (10 x 9 x 8)/(3 x 2) = 10 x 3 x 4 = 120.

Now, let’s examine the number of ways we can place the black balls in the row so that none of them are adjacent.

Case 1: If all the 3 black balls are somewhere between the white balls (i.e., none of black balls is first or last ball in the row of the 10 balls), then the number of ways is the same as the number of ways of writing the number 7 (the number of white balls) as a sum of 4 positive integers (for example, the arrangement WBWWWBWBWW is same as expressing 7 as 1 + 3 + 1 + 2). Let’s list them (note: we will first list 7 as a sum of 4 addends in ascending order and the total number of ways we can permute the 4 addends):

7 = 1 + 1 + 1 + 4 → Total number of ways = 4!/3! = 4

7 = 1 + 1 + 2 + 3 → Total number of ways = 4!/2! = 12

7 = 1 + 2 + 2 + 2 → Total number of ways = 4!/3! = 4

Therefore, the total number of ways 7 can be expressed as a sum of 4 positive integers is 4 + 12 + 4 = 20.

Case 2: If one of the 3 black balls is the first ball but neither of the other 2 black balls is the last ball of the row, then the number of ways is the same as the number of ways of writing 7 as a sum of 3 positive integers (for example, the arrangement BWWWWBWBWW is same as expressing 7 as 4 + 1 + 2). Let’s list them (using the same strategy as in case 1):

7 = 1 + 1 + 5 → Total number of ways = 3!/2! = 3

7 = 1 + 2 + 4 → Total number of ways = 3! = 6

7 = 1 + 3 + 3 → Total number of ways = 3!/2! = 3

7 = 2 + 2 + 3 → Total number of ways = 3!/2! = 3

Therefore, the total number of ways 7 can be expressed as a sum of 3 positive integers is 3 + 6 + 3 + 3 = 15.

Case 3: If one of the 3 blacks is the last ball but neither of the other 2 black balls is the first ball of the row, then the number of ways is the same as in case 2. That is, there are 15 ways this can happen.

Case 4: If a black ball is the first ball and another ball ball is the last ball of the row, then the number of ways is the same as the number of ways of writing 7 as a sum of 2 positive integers (for example, the arrangement BWWWWBWWWB is same as expressing 7 as 4 + 3). Let’s list them (using the same strategy as in case 1):

7 = 1 + 6 → Total number of ways = 2! = 2

7 = 2 + 5 → Total number of ways = 2! = 2

7 = 3 + 4 → Total number of ways = 2! = 2

Therefore, the total number of ways 7 can be expressed as a sum of 3 positive integers is 2 + 2 + 2 = 6.

Adding up the number of ways in the 4 cases above, the total number of ways the black balls in the row of 10 black and white balls such that none of the black balls are adjacent is 20 + 15 + 15 + 6 = 56. Therefore, the probability is 56/120 = 7/15.

Alternate Solution:

The number of ways 7 white balls and 3 black balls can be placed in a row is 10! / (7!3!) = (10 x 9 x 8)/(3 x 2) = 10 x 3 x 4 = 120.

Now, let’s examine the number of ways we can place the black balls in the row so that at least two of them are adjacent.

If all three black balls are adjacent, then we are arranging the objects [BBB]-W-W-W-W-W-W-W; i.e. 8 objects, 7 of which are indistinguishable. There are 8!/7! = 8 ways to arrange these objects.

If at least two black balls are adjacent, then we are arranging the objects [BB]-B-W-W-W-W-W-W-W; i.e. 9 objects, 7 of which are indistinguishable (notice that [BB] is considered as a single object, thus it can be distinguished from B). There are 9!/7! = 9 x 8 = 72 ways to arrange these objects.

Notice that the 72 ways of arrangements above actually include the ways where all three black balls are adjacent, since an arrangement can be [BB]-B-W-W-W-W-W-W-W. Further, each arrangement where all three black balls are adjacent is counted twice, as the above arrangement is counted again as B-[BB]-W-W-W-W-W-W-W. Thus, the number of arrangements where two black balls are adjacent but three black balls are not adjacent is 72 - 2 x 8 = 56.

Adding the arrangements where at most two and all three black balls are adjacent, we see that in 56 + 8 = 64 of the arrangements, at least two black balls are adjacent. Then, in 120 - 64 = 56 of the arrangements, no two black balls are adjacent. Since there are 120 total ways to arrange the balls and 56 of them are favorable, the probability that no two black balls are adjacent is 56/120 = 7/15.

Answer: B

Alternatively, instead of placing white balls first, we can place the black balls first:

_b_b_b_

Place a white ball between the B's. This leaves 5 white balls remaining to be placed anywhere in the 4 spots.

Using the sticks method, the black balls represent the sticks.

So 5 balls + 3 "sticks" = 8

8!/5!3!= 56

Ways to arrange all 10 balls without restrictions

10!/7!3! to take into account identical balls in the permutation =

120

Probability 56/120 = 7/15

Posted from my mobile device

Any quick solution to this problem that can ideally be done within 2 minutes?

One way to solve this one quickly

(I see that others have explained this method on this forum. But let me try to simplify the logic. Hopefully, it helps! This method takes less than a min if the concept is clear!)

1) Here are the 7 White Balls (visualize this).

O O O O O O O

2) Where can we put in the black balls? We have to put in the black balls in the spaces between the white balls, such that no two black balls are adjacent.

For instance:

B O O O B O O B O O is acceptable.

(1 black ball occupying the space before white ball 1, 1 occupying the space between white ball 3 and white ball 4, and the last occupying the space between white ball 5 and white ball 6)

3) What is not acceptable?

If more than one black ball occupies the space between any 2 white balls.

For instance:

O O B B O O B O O O is not acceptable. (2 black balls are occupying the space between white ball 2 and white ball 3)
O O O B B B O O O O is not acceptable. (3 black balls are occupying the space between white ball 3 and white ball 4)

4) So, in essence, how do we ensure that no black balls are adjacent?

We only need to ensure that not more than 1 black ball occupies any space between any 2 white balls.

5) How many spaces exist between the white balls? (Count the "_"s).

_O_O_O_O_O_O_O_ -> 8 spaces.

6) So, we need to ensure that the 3 black balls are placed somewhere in these 8 spaces, but we also have to ensure that no space gets more than 1 black ball.

How to do this?

By ensuring that we select 3 distinct spaces out of the 8 available spaces (for the 3 black balls).

In math terms -> 8C3 (selecting 3 distinct items out of 8 items).

So, actually, the number of arrangements where no 2 black balls are adjacent = 8C3 = 56.

Total number of arrangements possible between 7 white balls and 3 black balls = 10! / (7! 3!) = 120.

Our answer: 56/120 = 7/15. Choice B.

Hope this helps!

Harsha

how does distinct items equate to no black balls being adjacent help!

Hey!

If you meant "distinct spaces" rather than "distinct items", perhaps this will help ->

These are the white balls and the spaces between the white balls, where one or more black balls can be inserted->

_O_O_O_O_O_O_O_

Now, 3 black balls have to inserted somewhere above, such that no black balls are adjacent.

No black balls can be together. For instance, the following formations will not be allowed ->

_O B B O_O B O_O_O_O_

_O_O_O_O B B B O_O_O_

_O_O_O B B O_O_O_O B

Observe what happened in each of the above examples. More than 1 black ball got inserted into the same space.

1) 2 black balls got inserted into the space between White Ball (WB) 1 and White Ball 2.
2) 3 black balls got inserted into the space between WB4 and WB5.
3) 2 black balls got inserted into the space between WB3 and WB4.

So, what we don't want is for more than one black ball to occupy the spaces shown. So, we need to ensure that each of the 3 black balls occupy 3 different spaces among the 8 spaces shown. That is how we can ensure that there is no case of multiple black balls occupying the same space.

This is the logic behind choosing 3 separate spaces for the 3 black balls, from among 8 available spaces i.e., 8C3.

Hope this helps.
Harsha

1st approach -
I did the negation way -
so no adjacent will be = 1 - (2 together + 3 together)

2 together = lets take those 2B as one entity then (BB)WWWWWWWB so total arrangements=9!/7!*1!*1! (BB won't be arranged since same) = 56 ways
3 together = (BBB)WWWWWWW = 8!/7! = 8 ways
2together + 3together = 56+8 = 64 ways

now total ways = 10!/3!*7! = 120
so Ans = 1- (64/120) = 7/15

2nd approach (sticks and carrots) -
since no blacks adjacent so - _W_W_W_W_W_W_W_
so whites can only take these places (so 1 way) and Blacks can be arranged in these "_" so total ways = 8C3
ans = 1 * 8C3/10C3