Combinatorics DPP - RACE 2 - Clubbing and separation of objects - Easy
Q1 - In how many ways five different red balls and five different black balls can be arranged in a row such that,
(iii) no two red balls are together.
First arrange 5 black balls: 5!
Now there are 6 gaps for red balls: 6C5*5!
Answer: 5!.6C5.5! = 86400(It's the number of seconds in a day!)
(iv) All red balls are together and all black balls are together
2!.5!.5! = 2800
(iii) All vowels are not together
Total - (All vowels together) = 8!/2!2! - 720= 9360
(i) there is no restriction.
10!
10!
(ii) all red balls are not together.
Total: 10!
All red balls are together: 6!*5!
Answer: 10! - 6!5!
Total: 10!
All red balls are together: 6!*5!
Answer: 10! - 6!5!
(iii) no two red balls are together.
First arrange 5 black balls: 5!
Now there are 6 gaps for red balls: 6C5*5!
Answer: 5!.6C5.5! = 86400(It's the number of seconds in a day!)
(iv) All red balls are together and all black balls are together
2!.5!.5! = 2800
(v) Red and black ball are alternatively.
Arrange black balls: 5!
Now there are six gaps but we can choose either the first 5 or last 5 for the red balls: 2.5!
Answer: 2.5!.5! = 28800
Q2 -
Find the number of words which can be formed by using all letters 2A, E, I, 2B, C, D. If
Arrange black balls: 5!
Now there are six gaps but we can choose either the first 5 or last 5 for the red balls: 2.5!
Answer: 2.5!.5! = 28800
Q2 -
Find the number of words which can be formed by using all letters 2A, E, I, 2B, C, D. If
(i) Starts with C and end with A.
Total letters: 8
C------A
6!/2! = 360
Total letters: 8
C------A
6!/2! = 360
(ii) Vowels always occur together
(5!/2!)*(4!/2!) = 720
(5!/2!)*(4!/2!) = 720
(iii) All vowels are not together
Total - (All vowels together) = 8!/2!2! - 720= 9360
(iv) No two vowels are together
Remaining letters: 4
Arrange them: 4!/2!
Gaps: 5
Places for vowels: 5C4
Arrange: 4!/2!
Finally: 5.4!.4!/2!.2! = 720
Remaining letters: 4
Arrange them: 4!/2!
Gaps: 5
Places for vowels: 5C4
Arrange: 4!/2!
Finally: 5.4!.4!/2!.2! = 720
(v) Vowels & consonants are alternate
Vowels arrangement: 4!/2!
2 ways to choose 4 gaps out of 5.
Consonants arrangement: 4!/2!
Answer: 288
Vowels arrangement: 4!/2!
2 ways to choose 4 gaps out of 5.
Consonants arrangement: 4!/2!
Answer: 288
(vi) Vowels always occupy even places only
4!/2!.(4!/2!) = 144
4!/2!.(4!/2!) = 144
(vii) Order of vowels remains same.
Fix the vowels: AAEI
5C4.4!/2! for remaining = 60 (Can it be 840?)
Fix the vowels: AAEI
5C4.4!/2! for remaining = 60 (Can it be 840?)
(viii) Number of words formed by selecting 2 vowels and 3 consonants.
Vowels: 2A,E,I
Consonants: 2B,C,D
Choosing 2 Vowels: 2 Same or 2 distinct
Choosing 3 Consonants: 2 Same, 1 diff or all 3 diff
Case 1: 2 same vowels: 1C1
1a: 2 same consonants, 1 diff: 1C1.2C1
Arrangement: 5!/2!.2!
Answer: 60
1b: all 3 diff consonants: 3C3
Arrangement: 5!/2!
Answer: 60
Case 2: 2 distinct vowels: 3C2
1a: 2 same consonants, 1 diff: 1C1.2C1
Arrangement: 5!/2!
Answer: 360
1b: all 3 diff consonants: 3C3
Arrangement: 5!
Answer: 360
Total: 840
Vowels: 2A,E,I
Consonants: 2B,C,D
Choosing 2 Vowels: 2 Same or 2 distinct
Choosing 3 Consonants: 2 Same, 1 diff or all 3 diff
Case 1: 2 same vowels: 1C1
1a: 2 same consonants, 1 diff: 1C1.2C1
Arrangement: 5!/2!.2!
Answer: 60
1b: all 3 diff consonants: 3C3
Arrangement: 5!/2!
Answer: 60
Case 2: 2 distinct vowels: 3C2
1a: 2 same consonants, 1 diff: 1C1.2C1
Arrangement: 5!/2!
Answer: 360
1b: all 3 diff consonants: 3C3
Arrangement: 5!
Answer: 360
Total: 840
(ix) They contain the string 'EDBA' in them
Remaining letters: A,I,B,C
Arrange remaining: 4!
5 gaps to choose from: 5C1
Total: 5.4! = 120
Remaining letters: A,I,B,C
Arrange remaining: 4!
5 gaps to choose from: 5C1
Total: 5.4! = 120
Q3 - Let 'm' be the total number of words formed by arranging the letters of the word "SPECIAL" in all possible manner and 'n' be the number of words in which vowels occur alphabetically, then find m/n.
Solution:
m = 7!
A wrong way to compute n:
Arrange S,P,C,L: 4!
Choose 3 gaps out of 5: 5C3
Total: 5C3.4!
Why is this wrong?
Because this doesn't allow the vowels to sit together.
How to fix this?
If you want to continue using the gap idea, then use Stars and Bars.
Here n = 3, k =5. Why?
Recall that in stars and bars 'n' identical objects have to be put into 'k' distinct buckets.
So we use 'k-1' bars to divide the 'n' stars.
Here the vowels are effectively those objects because they are effectively identical(since ordering is not allowed, we can reframe it as ordering is immaterial so they are identical).
So (3+5-1)C(5-1) = 7C4
Multiply with 4! to get 7C4.4!
So m/n will be 6.
A simpler way to solve for 'n' is this:
Total ways: 7!
In this A,E,I will be present in 3! = 6 ways.
But we need only 1 way which is alphabetical. So divide and get 6.
Q4 - The number ways in which the letters of the word 'ARRANGE' be arranged so that
(ii) the two A's are together but not R's,
Q6 - In a G-20 summit, Mr. Trump wants to speak before Mr. Putin & Mr. Putin wants to speak after Mr. Modi. If the remaining speakers can speak at any number, then find the number of ways that all the 20 speakers can give their speeches.
Answer: 20!.2/3! = 20!/3
There are 6 ways to arrange T,M,P and there are only 2 valid:
T,M,P
T,P,M
M,P,T
M,T,P
P,T,M
P,M,T
Q7 - The total number of words formed using the letters of the word 'CHANDRASWAMI' which contains the word 'SCAM', is:
Total letters: 12
1C,1H,3A,1N,1D,1R,1S,1W,1M,1I
Solution:
m = 7!
A wrong way to compute n:
Arrange S,P,C,L: 4!
Choose 3 gaps out of 5: 5C3
Total: 5C3.4!
Why is this wrong?
Because this doesn't allow the vowels to sit together.
How to fix this?
If you want to continue using the gap idea, then use Stars and Bars.
Here n = 3, k =5. Why?
Recall that in stars and bars 'n' identical objects have to be put into 'k' distinct buckets.
So we use 'k-1' bars to divide the 'n' stars.
Here the vowels are effectively those objects because they are effectively identical(since ordering is not allowed, we can reframe it as ordering is immaterial so they are identical).
So (3+5-1)C(5-1) = 7C4
Multiply with 4! to get 7C4.4!
So m/n will be 6.
A simpler way to solve for 'n' is this:
Total ways: 7!
In this A,E,I will be present in 3! = 6 ways.
But we need only 1 way which is alphabetical. So divide and get 6.
Q4 - The number ways in which the letters of the word 'ARRANGE' be arranged so that
(i) the two R's are never together
Arrange the remaining: 2A, 1N, 1G, 1E : 5!/2!
6 gaps: choose 2: 6C2
Answer: 60*15 = 900
Arrange the remaining: 2A, 1N, 1G, 1E : 5!/2!
6 gaps: choose 2: 6C2
Answer: 60*15 = 900
(ii) the two A's are together but not R's,
Arrange the remaining: 1A(effectively), 1N, 1G, 1E : 4!
5 gaps: choose 2: 5C2
Answer: 24*10 = 240
5 gaps: choose 2: 5C2
Answer: 24*10 = 240
(iii) neither two A's nor two R's are together.
One wrong way to do it:
Remaining: N,G,E: 3!
Choose 2 gaps for A's: 4C2
Choose 2 gaps for R's: 6C2
Answer: 6*6*15 = 540
Why is it wrong?
Because it won't generate patterns like: ARANGRE since we never kept A's together in the first arrangement.
So, here is one correct way to do it:
Total ways: 7!/2!.2! = 1260
AA's together: 6!/2! = 360
RR's together: 360
AA's and RR's together: 5! = 120
Finally: 1260 - 360 - 360 + 120 = 660
One wrong way to do it:
Remaining: N,G,E: 3!
Choose 2 gaps for A's: 4C2
Choose 2 gaps for R's: 6C2
Answer: 6*6*15 = 540
Why is it wrong?
Because it won't generate patterns like: ARANGRE since we never kept A's together in the first arrangement.
So, here is one correct way to do it:
Total ways: 7!/2!.2! = 1260
AA's together: 6!/2! = 360
RR's together: 360
AA's and RR's together: 5! = 120
Finally: 1260 - 360 - 360 + 120 = 660
Q5 - In how many different ways can Indian cricket team of 11 players bat so that:
(i) Rohit wants to bat just before Kohli.
10!
10!
(ii) Rohit wants to bat before Kohli.
11!/2!
11!/2!
(iii) Rohit wants to bat before Kohli and Kohli wants to bat before Surya.
11!/3!Q6 - In a G-20 summit, Mr. Trump wants to speak before Mr. Putin & Mr. Putin wants to speak after Mr. Modi. If the remaining speakers can speak at any number, then find the number of ways that all the 20 speakers can give their speeches.
Answer: 20!.2/3! = 20!/3
There are 6 ways to arrange T,M,P and there are only 2 valid:
T,M,P
T,P,M
M,P,T
M,T,P
P,T,M
P,M,T
Q7 - The total number of words formed using the letters of the word 'CHANDRASWAMI' which contains the word 'SCAM', is:
Total letters: 12
1C,1H,3A,1N,1D,1R,1S,1W,1M,1I
Remove 4 and add 1 for 'SCAM'
Total letters: 9 (contains 2 A's)
9!/2!
Q8 - Number of permutations of 1,2,3,4,5,6,7,8 and 9 taken all at a time such that the digit 1 appearing somewhere to the left of 2, 3 appearing somewhere to the left of 4 and 5 somewhere to the left of 6, is:
9!/2!.2!.2!
Compare it to question 6, there the 3 letters were interlinked.
Here each pair is independent of others.
Another way:
Choose 2 places for first pair, second pair, third pair and arrange the remaining 3 numbers(7,8,9): 9C2.7C2.5C2.3!
Q9 - There are 10 white balls of different shades and 9 black balls of identical shades. Then the number of arrangements of these balls in a row such that no two black balls are together is:
One wrong way to do this:
Lay down 9 black balls: 1 way.
There are 10 gaps now.
10 distinct white balls in the gaps: 10!
Why is this wrong?
Because there are arrangements where we can put 2 white balls together and we are missing them in this method.
For e.g. we can do BW1,W2BW3...
So the better way to do it is this:
Firs arrange 10 white balls: 10!
Now choose 9 gaps from 11: 11C9
11C9.10! is the final answer.
Moral of the story: pick gaps for those objects which can't be put together and not vice versa.
Q10 - Three ladies have brought one child each for admission to a school. The principal wants to interview the six persons one by one subject to the condition that no mother is interviewed before her child. Then the number of ways in which interviews can be arranged are:
Solution:
This is similar to question 8:
6!/2!2!2! = 90
Why?
Total ways: 6!
Half of these will have M1 before C1. Half of remaining will have M2 before C2. Same for third child.
Another way to solve:
First choose 2 places for first Mother child pair: 6C2. And they can't interchange their places.
Then for next pair: 4C2
Then 2C2
Total: 6C2.4C2.2C2
Q11 - The number of permutations that can be formed by all the letters of the word 'DISTINCTION' in which all 3 I's appear before 'O', is:
Total letters: 11
1D, 3I, 1S, 2T, 2N, 1C, 1O
Total letters: 9 (contains 2 A's)
9!/2!
Q8 - Number of permutations of 1,2,3,4,5,6,7,8 and 9 taken all at a time such that the digit 1 appearing somewhere to the left of 2, 3 appearing somewhere to the left of 4 and 5 somewhere to the left of 6, is:
9!/2!.2!.2!
Compare it to question 6, there the 3 letters were interlinked.
Here each pair is independent of others.
Another way:
Choose 2 places for first pair, second pair, third pair and arrange the remaining 3 numbers(7,8,9): 9C2.7C2.5C2.3!
Q9 - There are 10 white balls of different shades and 9 black balls of identical shades. Then the number of arrangements of these balls in a row such that no two black balls are together is:
One wrong way to do this:
Lay down 9 black balls: 1 way.
There are 10 gaps now.
10 distinct white balls in the gaps: 10!
Why is this wrong?
Because there are arrangements where we can put 2 white balls together and we are missing them in this method.
For e.g. we can do BW1,W2BW3...
So the better way to do it is this:
Firs arrange 10 white balls: 10!
Now choose 9 gaps from 11: 11C9
11C9.10! is the final answer.
Moral of the story: pick gaps for those objects which can't be put together and not vice versa.
Q10 - Three ladies have brought one child each for admission to a school. The principal wants to interview the six persons one by one subject to the condition that no mother is interviewed before her child. Then the number of ways in which interviews can be arranged are:
Solution:
This is similar to question 8:
6!/2!2!2! = 90
Why?
Total ways: 6!
Half of these will have M1 before C1. Half of remaining will have M2 before C2. Same for third child.
Another way to solve:
First choose 2 places for first Mother child pair: 6C2. And they can't interchange their places.
Then for next pair: 4C2
Then 2C2
Total: 6C2.4C2.2C2
Q11 - The number of permutations that can be formed by all the letters of the word 'DISTINCTION' in which all 3 I's appear before 'O', is:
Total letters: 11
1D, 3I, 1S, 2T, 2N, 1C, 1O
Total ways: 11!/3!2!2!
3I and 1O can be arranged in 4!/3! = 4 ways.
But we just need 1 of those ways.
So we divide Total by 4.
Answer: 11!/3!2!2!.4 can be rewritten as 11!/4!2!2!
Q12 - There are 2 identical white balls, 3 identical red balls and 4 green balls of different shades. The number of ways in which these balls can be arranged in a row so that at least one ball is separated from the balls of same color, is:
Solution:
2W,3R,G1,G2,G3,G4
Total ways: 9!/2!3!
Same colors ball together: Treat W and R as 1. All G's together. 3!*4!
Answer: 9!/2!3! - 4!3! = 9.8.7!/12 - 4!3! = 6(7! - 4!)
Q13 - Number of all possible words that can be formed using all letters of the word 'MATHEMATICS' so that two M's and two T's are together but 2A's are not together:
Total letter: 11
2M,2A,2T,1H,1E,1I,1C,1S
First ignore A's and treat 2M and 2T as single letter: 7!
Put 2 A's in 8 gaps: 8C2
7!.8C2 is the final answer.
3I and 1O can be arranged in 4!/3! = 4 ways.
But we just need 1 of those ways.
So we divide Total by 4.
Answer: 11!/3!2!2!.4 can be rewritten as 11!/4!2!2!
Q12 - There are 2 identical white balls, 3 identical red balls and 4 green balls of different shades. The number of ways in which these balls can be arranged in a row so that at least one ball is separated from the balls of same color, is:
Solution:
2W,3R,G1,G2,G3,G4
Total ways: 9!/2!3!
Same colors ball together: Treat W and R as 1. All G's together. 3!*4!
Answer: 9!/2!3! - 4!3! = 9.8.7!/12 - 4!3! = 6(7! - 4!)
Q13 - Number of all possible words that can be formed using all letters of the word 'MATHEMATICS' so that two M's and two T's are together but 2A's are not together:
Total letter: 11
2M,2A,2T,1H,1E,1I,1C,1S
First ignore A's and treat 2M and 2T as single letter: 7!
Put 2 A's in 8 gaps: 8C2
7!.8C2 is the final answer.
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