Latest Elitmus Aptitude Question SOLUTION: Boxes numbered 1, 2, 3, 4 and 5 are kept in a row and they which are to be filled with either a red or a blue ball, such that no two adjacent boxes can be filled with blue balls. T

(d) 22
total number arrangement without any restriction is 2^5=32
nymber of arrangement in which blue balls are adjacent =bb,bb,bb,bb,bbb,bbb,bbb,bbbb,bbbb,bbbbb=10

hence required number= 32-10=22

I think none of these it should be 13
total ways=32 ,
Required arrangement= Total ways- (5 blue+ 4 blue+3 blue+2 blue)
=32-(1+5+9+4)=13

among them this 13 are following the condition i.e
1. RRRRR
2. BRRRR
3. RBRRR
4. RRBRR
5. RRRBR
6. RRRRB
7. BRRRB
8. BRRBR
9. BRBRR
10. RBRRB
11. RBRBR
12. RRBRB
13. BRBRB

in all other cases i.e 32-13=19 either two or three or four or all 5 balls are adjacent
1.BBBBB
2.BBBBR
3.BBBRB
4.BBRBB
5.BRBBB
6.RBBBB
7.RRBBB
8.RBRBB
9.RBBRB
10.RBBBR
11.BRRBB
12.BRBBR
13.BBRRB
14.BBRBR
15.BBBRR
16.RRRBB
17.RRBBR
18.RBBRR
19.BBRRR

Total number of ways filling the 5 boxes numbered as (1, 2, 3, 4 and 5) with either blue or red balls = 25 = 32
Two adjacent boxes with blue balls can be obtained in 4 ways, i.e., (12), (23), (34) and (45). Three adjacent boxes with blue balls can be obtained in 3 ways i.e., (123), (234) and (345). Four adjacent boxes with blue balls can be obtained in 2 ways i.e., (1234) and (2345) and five boxes with blue balls can be got in 1 way... so 32-10=22

First we find out all possible ways of filling 5 boxes, each with blue or red balls, no other restrictions. 5 boxes can be filled with balls of either of 2 colors in 2^5 ways, i.e. 32. i.e. The first box can be filled with either blue or red balls, thus can be filled in 2 ways. Same is the case for each of the other 4 boxes. Thus total number of ways is 2 * 2 * 2 * 2 * 2 = 32.

Then we figure out all the arrangements where two blue colored balls filled boxes are adjacent to each other. In this example all blue balls are identical to each other and all red balls are identical to each other. Thus the arrangement where boxes 1,2,3 have blue colored balls is same as the arrangement where boxes 3,2,1 have blue colored balls. This can happen in the following cases

All 5 boxes are filled with blue colored balls - This can happen in 1 way
4 boxes are filled with blue colored balls - This can happen as (1234, 1245, 1345, 2345), thus 4 ways
3 boxes are filled with blue colored balls - The possible arrangements are (123, 124, 125, 134, 145, 234, 245, 345), thus 8 ways. 135 is not counted as we need to have two adjacent boxes with blue colored balls.
2 boxes are filled with blue colored balls - The possible arrangements are (12, 23, 34, 45), thus 4 ways.
Thus, the total number of arrangements, where two boxes with blue colored balls are adjacent to each other, is 1 + 4 + 8 + 4 = 17 ways.

To get the desired answer, the number of arrangements where 5 boxes need to be filled, each with either blue or red colored balls, where no two adjacent boxes have blue colored balls, we subtract the number of arrangements where two blue colored balls filled boxes are next to each other, from the total number of possible arrangements.

No. of Arrangements (no two adjacent boxes have blue colored balls) = No. of Arrangements (Total possible) - No. of Arrangements (Blue colored balls in adjacent boxes)

Desired answer = 32 - 17 = 15.

tanvi jain is correct

total ways : 2^5 = 32 ,
Total ways - (2blues+3blues+4blues+5blues) together = 32 - (4+3+2+1) = 22
* Can use : number of permutations whn p types are alike = n!/p! *

Simply arrange blue balls with the given condition in 5 boxes.
0 blue balls - 1 way
1 blue ball - 5 ways
2 blue balls - 6 ways
3 blue balls - 1 way
4 blue balls - 0 way (why?)
add them to get 13 as the answer.

SIPRASTHITI MOHANTY is correct


http://www.pagalguy.com/articles/quantitative-aptitude-quiz-for-mba-entrance-exams-34897122

(b)10
since no two adjacent boxes can be filled with blue balls so we are left with the options as RR RB BR
BB can't be possible so three ways and there are 5 places so 5C3=10

First we find out all possible ways of filling 5 boxes, each with blue or red balls, no other restrictions. 5 boxes can be filled with balls of either of 2 colors in 2^5 ways, i.e. 32. i.e. The first box can be filled with either blue or red balls, thus can be filled in 2 ways. Same is the case for each of the other 4 boxes. Thus total number of ways is 2 * 2 * 2 * 2 * 2 = 32.

Then we figure out all the arrangements where two blue colored balls filled boxes are adjacent to each other. In this example all blue balls are identical to each other and all red balls are identical to each other. Thus the arrangement where boxes 1,2,3 have blue colored balls is same as the arrangement where boxes 3,2,1 have blue colored balls. This can happen in the following cases

All 5 boxes are filled with blue colored balls - This can happen in 1 way
4 boxes are filled with blue colored balls - This can happen as (1234, 1245, 1345, 2345), thus 4 ways
3 boxes are filled with blue colored balls - The possible arrangements are (123, 124, 125, 134, 145, 234, 245, 345), thus 8 ways. 135 is not counted as we need to have two adjacent boxes with blue colored balls.
2 boxes are filled with blue colored balls - The possible arrangements are (12, 23, 34, 45), thus 4 ways.
Thus, the total number of arrangements, where two boxes with blue colored balls are adjacent to each other, is 1 + 4 + 8 + 4 = 17 ways.

To get the desired answer, the number of arrangements where 5 boxes need to be filled, each with either blue or red colored balls, where no two adjacent boxes have blue colored balls, we subtract the number of arrangements where two blue colored balls filled boxes are next to each other, from the total number of possible arrangements.

No. of Arrangements (no two adjacent boxes have blue colored balls) = No. of Arrangements (Total possible) - No. of Arrangements (Blue colored balls in adjacent boxes)

Desired answer = 32 - 17 = 15.

The blue and red balls combination can be like this

red | blue
5 | 0
4 | 1
3 | 2
2 | 3
1 | 4
0 | 5

case1) 5 red----RRRRR 1 way

case2) 4 red and 1 blue
_ _ _ _ _ in those 5 places blue can be filled any one=> 5 ways and red can be in 1 way total=5 ways

case3) 3 red and 2 blue
_ _ _ _ _ blues shouldn't be adjacent so total ways is 5c2-4 (4 for adjacent)=6 ways and 3 red in 1 way
Total= 6*1 =6 ways

case4) 2red and 3 blue
_ _ _ _ _ only one way (1st place 3rd place 5th place)
Total=1 way

case5) 2 red and 4 blue
4 blue can't be placed such that no 2 are adjacent Total ways=0

case6) 5 blue
0 ways

Total=case1+case2+case3+case4+case5+case6
=1+5+6+1+0+0
=13 ways

i think the answer is 14.

total possible cases= 2^5= 32.

cases with adjacent blue balls:
case1(5 blue balls):
bbbbb-> 1

case2(4 blue balls):
rbbbb
brbbb
bbrbb
bbbrb
bbbbr

total of 5 combinations.

case3( 3 blue balls):

rrbbb
bbbrr
rbrbb
bbrbr
brrbb
bbrrb
rbbrb
brbbr

total of 6 combinations.

case 4( 2 blue balls):
rrrbb
bbrrr
rrbbr
rbbrr
total of 4 cases.

total cases for blue balls adjacent to one another: 1+5+8+4= 18

so, total combinations with no blue ball adjacent to one another is= 32-18=14.

_ * _ * _ * _
ans:
places for black boxes represented by _.
total ways: 5c5 (5 red balls) + 5c4.1c1 (4 red and 1 black) + 4c2.3c3 (2 black 3 red) + 3c3. 2c2 (3 black 2 red)
=13

22 is the right ans

Total number of ways of filling the 5 boxes numbered as (1, 2, 3, 4 and 5) with either blue or red balls = 25 = 32.

Now, let us determine the number of ways of filling the boxes such that the adjacent boxes are filled with blue.

If we decide to have 2 adjacent boxes with blue, it can be done in 4 ways, viz. (12), (23), (34) and (45).

If we decide to have 3 adjacent boxes filled with blue, it can be done in 3 ways, viz. (123), (234) and (345).

If we decide to have 4 adjacent boxes filled with blue, it can be done in 2 ways, viz. (1234) and (2345).

And all 5 boxes can have blue in only 1 way.

Hence, the total number of ways of filling the boxes such that adjacent boxes have blue = (4 + 3 + 2 +1) = 10.

Hence, the number of ways of filling up the boxes such that no two adjacent boxes have blue = 32 - 10 = 22

all red: _ _ _ _ _ = 1 combination;
4 red : _ _ _ _ = 5 combination (5 places to put 1 blue ball);
3 red : _ _ _ = 6 combination (4 places to put 2 blue ball);
2 red : _ _ = 1 combination (3 places to put 3 blue ball);
1 red : _ not possible;
Ans: 13

Total number of ways filling the 5 boxes numbered as (1, 2, 3, 4 and 5) with either blue or red balls = 25 = 32
Two adjacent boxes with blue balls can be obtained in 4 ways, i.e., (12), (23), (34) and (45). Three adjacent boxes with blue balls can be obtained in 3 ways i.e., (123), (234) and (345). Four adjacent boxes with blue balls can be obtained in 2 ways i.e., (1234) and (2345) and five boxes with blue balls can be got in 1 way.

Hence, the total number of ways of filling the boxes such that adjacent boxes have blue balls
= (4 + 3 + 2 + 1)
= 10

Hence, the number of ways of filling up the boxes such that no two adjacent boxes have blue balls
= 32 - 10
= 22

Total number of ways filling the 5 boxes numbered as (1, 2, 3, 4 and 5) with either blue or red balls = 25 = 32
Two adjacent boxes with blue balls can be obtained in 4 ways, i.e., (12), (23), (34) and (45). Three adjacent boxes with blue balls can be obtained in 3 ways i.e., (123), (234) and (345). Four adjacent boxes with blue balls can be obtained in 2 ways i.e., (1234) and (2345) and five boxes with blue balls can be got in 1 way.

Hence, the total number of ways of filling the boxes such that adjacent boxes have blue balls
= (4 + 3 + 2 + 1)
= 10

Hence, the number of ways of filling up the boxes such that no two adjacent boxes have blue balls
= 32 - 10
= 22