Combinatorics DPP - RACE 6 - Q16 pending discussion
Q1. A question paper consists of two sections. Section A has 7 questions and section B has 8 questions. A student has to answer 10 questions out of these 15 questions. The number of ways in which he can answer if he must answer at least 3 of section A and at least 4 of the section B is:
Solution:
One wrong way to solve this:
For the mandatory questions: 7C3*8C4
Now 3 more questions need to be answered from 4 of A and 4 of B.
Cases: 3A,0B | 2A,1B | 1A,2B | 0A,3B
4C3*4C0 + 4C2*4C1 + 4C1*4C2 + 4C0*4C1
= 4 + 24 + 24 + 4 = 56
Finally: 7C3*8C4*56 = 35*70*56 = 137200
Why is it wrong?
Let's say we chose A1,A2,A3 as mandatory from A and then chose A4,A5,A6 when we did 4C3*4C0 for (3A,0B).
But we could have done it the other way round also, i.e. choose A4,A5,A6 as mandatory and then choose A1,A2,A3. So there is overcounting.
Correct Solution:
Simply create the cases for A,B:
(3,7),(4,6),(5,5),(6,4)
7C3*8C7 + 7C4*8C6 + 7C5*8C5 + 7C6*8C4 = 2926.
Q2. A kindergarten teacher has 25 kids in her class. She takes 5 of them at a time, to a museum as often as she can, without taking the same 5 kids more than once. Then the number of visits, the teacher makes to the museum exceeds that of a kid by:
Solution:
Total trips for the teacher: 25C5 = 25.24.23.22.21/5.4.3.2 = 5*23*22*21
For the kid: 24C4 = 24.23.22.21/4.3.2 = 23.22.21
Difference = 4*23*22*21 = 42504
Or by Pascal's rule:
25C5 - 24C4 = 24C5
Q3. In an examination, there are 4 compulsory and 3 optional subjects. A student has to appear in all 7 papers and to pass the examination he must pass all the 4 compulsory subjects and at least two optional subjects. The number of ways in which he can fail is:
Solution:
Each exam has 2 outcomes: Pass/Fail
Total possible outcomes: 2^7 = 128
Number of ways to pass: All 4 compulsory subjects should be 'P' and 2/3 out of optional should be 'P'.
So final successful strings will be :
PPPP.PPF(4 total variants, PFP, FPP, PPP, PPF)
Number of ways to fail = 128 - 4 = 124
Q4. Find total number of words formed using the letters of the word "APRAJITA"
Total letters: 8
3A,1I,1P,1R,1J,1T
Total ways to arrange: 8!/3! = 6720
(A) If all three A's appear before I:
AAAI can occur in 4 variations. We need one of them. so 6720/4 = 1680
(B) If vowels and consonants are alternate:
There are 4 vowels and 4 consonants.
So a word either starts from a vowel and alternates or with consonant and alternates.
So 2.(4!/3!).4! = 192
(C) If relative position of vowels and consonants remain same:
What this means is that vowels will remain at positions 1,4,6,8 and consonants at 2,3,5,7
Total possibilities of vowel ordering: 4!/3! = 4
Consonants will appear in 4! ways.
Total: 4.4! = 96
(D) If two A's are together but separated from third A.
Arrange 5 remaining letters: 5!.
Put AA and A in 2 out of 6 gaps: 6C2 = 15
Permute AA and A: 2!
Answer: 120*15*2 = 3600
Q5. The number of ways in which 8 balls of different colors can be arranged in a row so that the two balls of particular colors (say red and white) may never come together is:
Solution:
Arrange the remaining 6 balls: 6!
Choose 2 gaps: 7C2
Permute 2 balls: 2!
6!.7C2.2! = 30420
Q6. If repetition of digits are not allowed, then find total number of 7 digit numbers that can be formed using the non-zero digits:
(A) if product of any 5 consecutive digits is divisible by 5:
5 has to be there at position 3,4 or 5. So 3C1 places for 5.
Remaining 6 digits will be chosen and arranged: 8C6*6!
Answer: 8!.3/2! = 12.7!
(B) if product of all digits of the number is divisible by 12:
12 = 2.2.3
So we need two 2s and one 3.
Let's group the numbers by what they contribute:
Group 1 => 1,5,7 => Neither 2 nor 3
Group 2 => 2,4,8 => Contribute 2's
Group 3 => 3,9 => Contribute 3's
Group 4 => 6 => Contribute both 2 and 3.
If we start enumerating all the ways to get 12, it will be quite large.
So let's find the ways when it's not divisible by 12.
Out of 7, first choose 3 numbers from Group 1. Now 4 digits remaining.
If we choose all 3 from Group 2, we still need to choose 1 from Group 3 or 4. Which will make it divisible by 12.
If we choose all 3 from Group 3,4, we still need to choose one from Group 2. Which will make it divisible by 12.
So there is no way to get a number which is not divisible by 12.
So, all numbers will be divisible by 12.
Answer: 9C7.7! = 36.7!
(C) if number is divisible by 3:
Group the digits by the remainder by 3:
0 => 3,6,9
1 => 1,4,7
2 => 2,5,8
Since we have to select 7 digits, we have to include at least one digit from each group.
Case 1: one digit from '0' group, number will be like 1112220 => it's divisible by 3 => 7!.3C1 ways
Case 2: two digits from '0' group, 0011222 or 0022111 => not divisible by 3
Case 3: three digits from '0' group, 0001122 or 0001112 or 0001222, only 0001122 is div by 3 => 3C2.3C2.7! => 9.7!
Total: 12.7! ways
Q7. A number is called a palindrome if it reads the same backward as well a forward. For example 285582 is a six digit palindrome. If the sequence of all positive palindromes are written in ascending order:
1,2,3,4,5,6,7,8,9,11,22......
Then find
(A) the 2018th positive palindrome:
Enumerate:
a => 9
aa => 9 (18)
aba => 9*10 = 90 (108)
abba => 9*10 = 90 (198)
abcba => 9*10*10 = 900 (1098)
abccba => 9*10*10 = 900 (1998)
Now 20 remaining:
Pattern would be abcdcba
100.0.001 to 100.9.001 (2008)
101.0.101 to 101.9.101 (2018)
Answer: 1019101
(B) the rank of 2140412:
Till 6 digits we had got 1998.
A simple way to count all 7 digit palindromes till the given number = 2140 - 1000 + 1 = 1141
Answer: 1998 + 1141 = 3139
Now the pattern is:
abc.d.cba
a = 1 => 10*10*10 = 1000 numbers (2998)
a = 2, b = 0 => 10*10 = 100 ( 3098)
a = 2, b = 1, c = 0 to 3 => 4*10 = 40 (3138)
Last number so far is 213.9.312
Next number is 214.0.412 (3139)
Answer: 3139
Q8. Let each side of smallest square of chess board is one unit in length. Then, find
(A) The total number of rectangles on the chess board with one side even and one side odd:
Remember from an earlier question that there are 9C2*9C2 ways of making rectangles on an 8x8 chessboard. Because there are 9 horizontal and 9 vertical lines. And we need to choose 2 from each.
Now we have to choose lines in a way that their gap is odd or even.
Let's say we choose vertical lines in a way that their gap is odd:
Line 1 can go with lines 2,4,6,8
2 => 3,5,7,9
Similarly 3,4 will have 3 options each.
5,6 will have 2 options each.
7,8 will have 1 option each.
Total: 8 + 6 + 4 + 2 = 20
Now we choose horizontal lines in a way that their gap is even:
1 => 3,5,7,9
2 => 4,6,8
3 => 5,7,9
4,5 will get 2 each
6,7 will get 1 each.
8,9 will get 0 each.
Total: 4 + 6 + 4 + 2 = 16
Now same thing can be done with vertical even and horizontal odd.
So final answer: 2*16*20 = 640
(B) The total number of rectangles on the chess board whose area is odd:
Both sides should be odd: 20*20 = 400
(C) The number of ways in which a pair of smallest square can be selected which has exactly one corner common:
Start from the first row: 1st square and the 8th square can be paired with 1 square from the row below. And the remaining 6 can be paired with 2. So total: 1 + 1 + 12 =14.
Now same number of ways to create pairs from 2nd and 3rd row and so on till 7th/8th row.
So 14*7 = 98.
Q9. Let a person has to go from A to D moving along horizontal and vertical grids using shortest path (as shown in figure). Then find:
He can go via M or N.
To reach M: 5U,2R => 7!/5!2! = 21
To reach N: 4U,2R => 6!/4!2! = 15
From M to D: 2R,2U => 4!/2!2! = 6
From N to D: 2R,3U => 5!/2!3! = 10
A -> M -> D = 21*6 = 126
A -> N -> D = 15*10 = 150
But there is a corner case here. You can reach M via N.
A->N->M->D
15*1*6 = 90
These paths will be counted for both M,N.
So total ways: 126 + 150 - 90 = 186
126 - 90 = 36
(C) The number of paths in which he does not cross point M.
150 - 90 = 60
Q10. The number of ways in which 200 different things can be divided into groups of 100 pairs, is:
Solution:
For the first object there are 199 choices.
Then for the third there are 197.
...
Answer: 199.197...1
Generic solution for 2n objects:
(2n-1).(2n-3)...1
Multiply by missing numbers up and down.
2n.(2n-1).(2n-2)...2.1/2n.(2n-2)...2
(2n)!/2^n.n!
Or another way to see this:
Arrange objects in a line in (2n)! ways.
(A,B) and (B,A) are same. So divide by 2^n to account for each pair.
(2n)!/2^n
Now each pair arrangement is overcounted by n!, since (A,B),(C,D) and (C,D),(A,B) are practically same. So divide by n!.
(2n)!/2^n.n!
is the final answer.
Also, straightforward application of grouping theory.
Q11. Five balls are to be placed in three boxes. Then find the number of different ways these balls can be placed so that no box remains empty, if
(A) balls and boxes are different:
Ways to distribute:
1,2,2 => 5!/1!2!2!2! = 15
1,1,3 => 5!/1!1!2!3! = 10
Then 3! for arranging in boxes.
Total 25*6 = 150
(B) balls are identical and boxes are different
Stars and bars.
k = 3
Give 1 to each, remaining n = 2
(2+3-1)C(3-1) = 4C2 = 6
(C) balls are different and boxes are identical
This is same as (A) but without 3! for boxes.
So: 25.
(D) balls as well as boxes are identical
2 ways to make groups so answer is 2.
Q12. Let N = 2^3.3^5.5^7, then find
(A) Number of divisors of N which are divisible by 3 but not by 12:
2 ways to select 2.
5 ways to select 3.
8 ways to select 5.
Total: 80
(B) Sum of all divisors of N:
(2^0 + 2^1 ...2^3)(3^0+..+3^5)(5^0.+..+5^7)
= (2^4 - 1)/(2-1) * (3^6 - 1)/(3-1) * (5^8 - 1)/(5-1)
= (2^4 - 1) * (3^6 - 1) * (5^8 - 1)/8
This is by sum of geometric series formula.
(C) Sum of reciprocal of all divisors of N:
Now you will multiply 3 different geometric series sums same as (B). But now common ratio will be 1/2,1/3,1/5.
You will get:
(2^4 - 1) * (3^6 - 1) * (5^8 - 1)/8(2^3.3^5.5^7 = N)
Q13. There are 10 seats in a row of which 4 are to be occupied. The number of ways of arranging 4 persons so that no two persons sit side by side, is:
Solution:
This problem can be rephrased as this:
6 people are sitting already and can't move. We have to put 4 people in the gaps in and around them.
Total gaps: 7
7C4 ways to choose. 4! to arrange.
7C4*4! = 840.
Q14. Total number of positive integral solutions of the inequation 10 < x + y + z <= 25 is:
Solution:
x + y + z <= 25
x + y + z + w = 25
x' + y' + z' + w = 22
(22+4-1)C(4-1) = 25C3 total solutions.
25C3 = 25.24.23/6
x + y + z <= 10
x + y + z + w = 10
x' + y' + z' + w = 7
(7+4-1)C(4-1) = 10C3
10C3 = 10.9.8/6
25C3 - 10C3 = 1/6(25.24.23 - 10.9.8) = 8.5.3.(5.23 - 6)/6 = 20(109) = 2180
Then find the number of different ways they can dance if :
D0 = 1, D1 = 0, D2 = 1, D3 = 2, D4 = 9, D5 = 44
(A) No husband dances with his own wife.
!5 = D5 = 44
(B) Exactly 2 husbands dance with their respective wives.
5C2*D3 = 10*2 = 20
(C) At least 2 husbands do not dance with their respective wives.
5C2*D2 + 5C3*D3 + 5C4*D4 + 5C5*D5 = 10 + 20 + 45 + 44 = 119
(D) At least 1 and at most 3 husbands dance with their respective wives.
5C2*D2 + 5C3*D3 + 5C4*D4 = 10 + 20 + 45 = 75
Q16. 5 students each gave their jacket & umbrella to the guard at the entrance of Reliable tower in order to attend the grand result celebration party at Reliable. At the end of the night, the guard was in a mood of enjoyment and gave the students a random jacket & a random umbrella in such a way that each person got a pair of jacket & umbrella in a random manner. Then find:
(A) The number of ways in which each student gets the wrong jacket and the wrong umbrella.
D5*D5 = 44*44 = 1936
(B) The number of ways in which no student gets both the right jacket and the right umbrella.
Solution:
First compute total possible assignments: 5! * 5! (independently for jackets and umbrellas) = 14400.
Now subtract the cases when at least one student got both right.
Let's say A1 is the set of assignments when one student gets both jacket and umbrella right.
There are 4! * 4! ways because remaining stuff can be assigned freely.
Total such assignments for 5 students = 5C1*(4!)^2 = |A1|
Similarly A2,A3,A4,A5 are assignments where 2,3,4,5 students got both the things right.
|A2| = 5C2*(3!)^2 and so on...
Now notice that A1,A2,A3,A4,A5 are overlapping sets.
Because when we assigned stuff to one student correctly, that might have included the cases when other students got it right as well.
So to compute the size correctly here we need to use principal of inclusion/exclusion for overlapping sets union.
So |A1 U A2 U A3 U A4 U A5| =
5C1.(4!)^2 - 5C2.(3!)^2 + 5C3.(2!)^2 - 5C4.(1!)^2 + 5C5.(0!)^2
= 2556
14400 - 2556 = 11844
A wrong solution:
In this solution we would add the following cases:
1. Every student gets both items wrong. D5*D5 = 44*44
2. Exactly one student gets exactly one correct item.
Let's jacket is that correct item.
Choose one student: 5C1
De-arrange the remaining 4 jackets: D4 = 9
De-arrange the 5 umbrellas: D5 = 44
Repeat the same with correct umbrella: 2C1
2C1 * 5C1 * D4 * D5 = 2C1 * 5C1 * 9 * 44
3. Exactly 2 students get exactly one correct item.
2C1 * 5C2 * D3 * D5 = 2C1 * 5C1 * 2 * 44
4. Exactly 3 students get exactly one correct item.
2C1 * 5C3 * D2 * D5 = 2C1 * 5C1 * 1 * 44
5. Exactly 5 students get exactly one correct item.
2C1 * 5C5 * D0 * D5 = 2C1 * 5C1 * 1 * 44
Note that exactly 4 students getting exactly one correct item is not possible because that would mean 5th student is also getting the correct item.
Consider the case 3. We only considered the cases when each student receives 2 umbrellas correct or 2 jackets correct. What about the case when 1 jacket and 1 umbrella are correctly given out. Similarly for the subsequent cases.
So let's fix this wrong solution.
a. The number of ways in which each student gets the wrong jacket and the wrong umbrella.
D5*D5 = 44*44 = 1936.
b. Exactly one student gets right jacket OR right umbrella:
Choose jacket OR umbrella: 2C1
Choose one student: 5C1
Derange 4 items = D4 = 9
Derange 5 items = D5 = 44
2C1*5C1*9*44 = 3960
c. exactly 2 students get right jacket OR right umbrella:
2 right jacket + 2 right umbrella = 2C1 * 5C2 * D3 * D5 = 1760
1 right jacket + 1 right umbrella = 5C1 * 4C1 * D4 * D4 = 1620
Total: 3380
d. Exactly three students get right jacket OR right umbrella:
3 right umbrellas + 3 right jackets = 2C1*5C3*D2*44 = 880
2 Umbrella 1 jacket + 2 Jacket 1 umbrella = 2C1*5C2*3C1*D3*D4 = 2*10*3*2*9 = 1080
Total: 880 + 1080 = 1960
e. Exactly 4 students get right jacket OR right umbrella:
3 jacket 1 umbrella or vice versa = 2C1 * 5C3 * 2C1 * D2 * D4 = 2*10*2*1*9 = 360
2 jacket 2 umbrella = 5C2 * 3C2 * D3 * D3 = 10*3*2*2 = 120
Total: 480
f. Exactly five students get right jacket OR right umbrella:
5 right jackets + 5 right umbrellas = 2C1*5C5*D0*44 = 88
4 right jackets and 1 right umbrella(or vice versa): impossible coz 4 right jackets mean 5 right jackets.
3 right jackets and 2 right umbrellas + vice versa = 2C1 *5C2*D2*D3 = 2*10*2 = 40
Total: 88 + 40 = 128.
Final count: 1936 + 3960 + 3380 + 1960 + 480 + 128 = 11844
(C) The number of ways in which each student gets at least one of his own jacket or his own umbrella.
Solution:
Total ways - (at least one student gets both jacket/umbrella wrong) = A - B
A = 5!.5! = 14400
B = B1 U B2 U B3 U B4 U B5
B1 = one student gets both wrong
B2 = two students get both wrong
...
B5 = five students get both wrong
|B| = |B1| - |B2| + |B3| - |B4| + |B5|
=
For B1, we first compute the jacket part and square it since umbrella part is also same and independent.
One student gets the jacket wrong = Total ways of arranging jackets(5!) - he gets the jacket right(4! to arrange other jackets) = 120 - 24 = 96.
Can be done in 5C1 ways.
Same for umbrella.
B1 = 5C1*(96^2)
For B2:
Two student gets the jacket wrong =
Total ways of arranging jackets(5!) - two students get the jacket right
= 120 - (24 + 24 - 3!) = 78
24 + 24 - 3! = First student gets the jacket right + Second also - Both get right (Inclusion/Exclusion)
= 2C1.4! - 2C2.3!
B2 = 5C2*(78^2)
For B3:
Three student gets the jacket wrong =
120 - 3 students get right
3 students get right =
3C1.4! - 3C2.3! + 3C3.2! = 72 - 18 + 2 = 66
B3 = 5C3*(120-66)^2 = 5C3*(64^2)
For B4:
Four students get the jacket wrong =
120 - 4 students get right
4 students get right =
4C1.4! - 4C2.3! + 4C3.2! - 4C4.1! = 96 - 36 + 8 -1 = 67
B4 = 5C4*(120-67)^2 = 5C4*(53^2)
For B5:
Five students get the jacket wrong =
120 - 5 students get right
5 students get right =
5C1.4! - 5C2.3! + 5C3.2! - 5C4.1! + 5C5.0! = 120 - 60 + 20 -5 + 1 = 76
B5 = 5C5*(120-76)^2 = 5C5*(44^2)
Now:
Recall that:
|B| = |B1| - |B2| + |B3| - |B4| + |B5|
= 14091
Answer: Total ways - (at least one student gets both jacket/umbrella wrong) = A - B
A = 5!.5! = 14400
A - B = 14400 - 14091 = 309 is the answer.
Q17. A cat is trapped at the bottom of a well having N stairs to climb up. In each step, the cat can cover one, two or three stairs at a time. Then the number of different ways in which the cat can come out are if N = 6, 9, 10.
Solution 1:
For stair numbered N you can reach it via stair numbers N-1,N-2,N-3.
For example look at this picture to figure out how many ways are there to reach stair number 3.
So we can say that:
f(n) = f(n-1) + f(n-2) + f(n-3)
Which means that add the number of ways to reach the stairs numbered n-1,n-2,n-3.
For the bases cases, note that:
f(1) = 1, f(2) = 2, f(0) = 1
Why is f(0) = 1?
Because there is 1 way to reach stair numbered 0(bottom of the well) - do nothing.
Or because f(3) = f(0) + f(1) + f(2) => 4 = f(0) + 1 + 2 => f(0) = 1
So f(4) = f(3) + f(2) +f(1) = 7
f(5) = 13
f(6) = 24
f(7) = 44
f(8) = 81
f(9) = 149
f(10) = 274
Solution 2:
a + 2b + 3c = 6
Ordered pairs:
(6,0,0) => 6 1's => Number of ways to arrange => 6!/6!.0!.0! = 1
a = 5 not valid
(4,1,0) 4 1's, 1 2's => Number of ways to arrange => 5!/4!1! = 5
(3,0,1) => 4!/3!1! = 4
(2,2,0) => 4!/2!2! = 6
(1,1,1) => 3! = 6
(0,0,2) => 1
(0,3,0) => 1
Total: 24
Similarly you can do it for 9,10
Q18. If A = {1, 2, 3, 4 ....2023} and B,C are subsets of A
(A) Number of ways of selecting unordered pairs of B and C:
Number of subsets of A = 2^2023
Pick any 2 subsets and forget the order: nC2 = (2^2023)C2 = (2^2023)(2^2023-1)/2 = 2^2022(2^2023-1)
But this only considers the cases when B and C are unequal.
We need to add 2^2023 for the cases when they are equal.
So final answer: 2^2022.(2^2023 + 1)
(B) Number of ways of selecting unordered pair of B and C such that B intersect C = {}:
Each element can be in B or C or none of them.
So 3 possible states => Number of Ordered pairs = 3^2023
How to convert to unordered, i.e. {A,B} and {B,A} are same.
Note that here each pair except {phi, phi} will occur twice.
So (3^2023 - 1)/2 + 1 = (3^2023 + 1)/2
(C) Number of ways of selecting unordered pair of B and C such that B U C = A,B intersect C = {}
Ordered pairs => Each element will be in B or C so 2 states => 2^2023
Unordered pairs = (2^2023)/2 = 2^2022.
(D) Number of ways of selecting ordered pair of B and C such that B intersect C is singleton.
Singleton means a set with exactly one element.
So B and C will have exactly one element in common.
Let's say 1 is common in B,C.
Rest of the elements can either be in B or in C or in none of them.
So 3 possible states for each remaining element => 3^2022
And this will repeat 2023 times.
So 2023.3^2022
Q19. 16 players P1, P2, P3,....,P16 take part in a tennis tournament. Lower suffix player is better than any higher suffix player. These players arc to be divided into 4 groups each comprising of 4 players and the best from each group is selected for semifinals.
(A) Number of ways in which they can be divided into 4 equal groups if players P1, P2, P3 and P4 are in different groups, is:
Remaining players: 12
4 groups of 3 players each => 12!/(3!^4).4!
Now assign these groups to P1,P2,P3,P4 => 4!
Answer: 12!/(3!^4) = 11!/108
(B) Number of ways in which these 16 players can be divided into four equal groups such that when the best player is selected from each group, P6 is one among them, is:
So P6 can't be in a group where P1,P2,P3,P4,P5 are present.
Solution:
Choose 3 players for P6: 10C3
For the remaining 12 players, create 3 groups of 4 each => 12!/(4!4!4!3!)
Multiply both => 10C3*12!/(4!4!4!3!) = 20.12!/(4!^3)
Another solution:
So let's find number of groups when P6 is present with these people and subtract from total.
P6 with 3 better players: 5C3 = 10
P6 with 2 better players: 5C2*10C1 = 100
P6 with 1 better player: 5C1*10C2 = 225
Total: 335
And the remaining 12 players to be arranged in 3 groups of 4 each => 12!/4!.4!.4!.3!
Total possible groups: 16!/{(4!^4).4!}
Answer: 16!/(4!^5) - 335*12!/(4!4!4!3!)
Q20. Consider the word W = BOOKKEEPING
(A) Total number of combinations of 5 letters taken from the letters of the word 'W' is:
Total letters: 11
1B, 2O, 2K, 2E, 1P, 1I, 1N, 1G
Case 1: All 5 distinct: 8C5 = 56
Case 2: 1,1,1,2 => 3C1*7C3 = 105
Case 3: 1,2,2 => 3C2*6C1 = 18
Total: 179
(B) The total number of words that can be formed using the letters of the word 'W' such that identical letters never occur together, is:
We will subtract from total those words which have identical letters together.
Total: 11!/2!2!2!
With OO (similarly for KK,EE): 10!/2!2!
With OO && KK (similarly for other OO && EE, EE && KK): 9!/2!
With OO && KK && EE: 8!
Now we need to use PIE:
Total cases when identical pairs occur together = 3.10!/2!2! - 3.9!/2! + 8!
Subtract from total: 11!/2!2!2! - 3.10!/2!2! + 3.9!/2! - 8!
= 7!(11.10.9 - 3.10.9.2 + 3.9.4 - 8) = 7!(450 + 100) = 7!.550
One wrong way to solve:
Arrange the 5 single letters: 5!
Now there are 6 gaps, choose 6: 6C6 = 1
Now arrange 6 letters(which have 2O,2K,2E): 6!/2!2!2! = 90
Answer: 5!*90 = 10800
Why is this wrong?
Because we are forbidding cases like 'KO', 'KE' etc.
Q21. A mega pizza is to be sliced n times and Sn denotes the maximum number of pieces after n slice:
(A) Relation between Sn and Sn–1 is:
S1 = 2
S2 = 4
S3 = 7
S4 = 11
S_n = Sn_1 + n
(B) If the mega pizza is to be distributed among 60 persons, each one of them get at least one piece, then minimum number of ways of slicing the mega pizza is:
Sn = n + n-1 + n-2....S1 = (n^2 + n + 2)/2 >= 50 => n = 11 is the least value.
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