Permutations and combinations Problems & Solutions Part-II

  After discussing the definitions and basic concepts we will enlist all the results and relations of permutation and combination, depending on all those we will get more familiar with the concept of permutation and combination by solving miscellaneous examples.

Points to remember (Permutation)

  1. The number of ways of ordering = nPr={n(n-1)(n-2)…..(n-r+1)((n-r)!)}/(n-r)!= n!/{(n-r)!}
  2. The number of arranging of n different objects taken all together at a time is = nPn =n!
  3. nP0 =n!/n!=1
  4. P=n. n-1Pr-1
  5. 0!=1
  6. 1/(-r)!=0 , (-r)!=∞ (r N)
  7. The number of ways of filling r places where every place can be filled by any one of n objects, The count of permutations = The number of ways of stuffing r places =(n)r   

Example: How many numbers between 999 and 10000 can be generated with the help of numbers 0, 2, 3,6,7,8 where the digits must not be duplicated?

Solution: The numbers in between 999 and 10000 all are of four digit numbers.

                   The four-digit numbers constructed by digits 0, 2, 3,6,7,8 are

Permutation
Permutation: Example

  But here those numbers are also involved which begin from 0. So we can take the numbers which are formed with three digits.

Taking initial digit 0, the number of ways to arrange pending 3 places from five digits 2, 3,6,7,8 are 5P3 =5!/(5-3)!=2!*3*4*5/2!= 60

So the required numbers = 360-60 = 300.

Example: How many books can be set out in a row so that the two books mentioned are not together?

Solution: Total number of orders of n different books =n!.                                                                                                                

           If two mentioned books always together then number of ways =(n-1)!x2

Example: How many ways are there divided by 10 balls between two boys, one getting two and the other eight.

Solution: A gets 2, B  gets 8;  10!/2!8!=45

                  A gets 8, B gets 2; 10!/(8!2!)=45

that means 45+45=90 ways the ball will be divided.

Example: Search the number of arranging of the alphabets of the word “CALCUTTA”.

Solution: Required number of ways =8!/(2!2!2!)=5040

Example: Twenty people have been invited to the party. How many different ways in which they and the host can sit at a round table, if the two people have to sit on either side of the keeper.

Solution: There will be total 20 + 1 = 21 persons in all.

The two specified persons and the host be considered as one unit so that these remain 21 –  3 + 1 = 19 persons to be arranged in 18 ! ways.

 But the two particular person on either side of the host can themselves be arranged in 2! ways.

  Hence there are 2 ! *18 ! ways.

Example : In how many ways a garland can be made from exactly 10 flowers.

Solution:  n flowers’ garland can be made in (n-1)! ways.

Using 10 flowers garland can be prepared in 9!/2 different ways.

Example: Find the specific four-digit number which should be formed by 0, 1, 2, 3, 4, 5, 6, 7 so that each and every number has the number 1.

Solution: After securing 1 at first position out of 4 places 3 places can be filled by7P3 =7!/(7-3)!=5*6*7=210 ways.

But some numbers whose fourth digit is zero, so such type of ways =6P2=6!/(6-2)!=20.

                   Total ways = 7P36P2 =210-20=180

Keep these Points in mind for Combination

  • The number of combinations of n objects, of which p are identical, taken r at a time is

n-pCr+n-pCr-1+n-pCr-2+……..+n-pC0 , if r<=p and  n-pCr+n-pCr-1+n-pCr-2+…..+n-pCr-p  , if r>p

  1. n choose 0 or n choose n is 1, nC0 = nCn =1, nC1 =n.
  2. nCr + nCr-1 = n+1Cr
  3. Cx = nCy <=> x=y or x+y=n
  4. n. n-1Cr-1 =(n-r+1) nCr-1
  5. nC0+nC2+nC4+….=nC1+nC3+nC5…..=2n-1
  6. 2n+1C0+2n+1C1+2n+1C2+……+2n+1Cn=22n
  7. nCn+n+1Cn+n+2Cn+n+3Cn+………..+2n-1Cn=2nCn+1
  8. Number of combinations of n dissimilar things taken all at a time. nCn=n!/{n!(n-n)!}=1/(0)!=1

In continuation we will solve some examples  

Example: If 15Cr=15Cr+5 , then what is the value of r?

Solution: Here we will use the above

 nCr=nCn-r on the left side of the equation

15Cr=15Cr+5 => 15C15-r =15Cr+5

=> 15-r=r+5 => 2r=10 => r=10/2=5

so the value of r is 5 implies the problem of 15 CHOOSE 5.

Example: If 2nC3 : nC2 =44:3   find the value of r, So that the value of nCr  will be 15.

 Solution: Here the given term is the ratio of 2n choose 3 and n choose 2 as

by the definition of combination

(2n)!/{(2n-3)!x3!} X {2!x(n-2)!}/n!=44/3

=> (2n)(2n-1)(2n-2)/{3n(n-1)}=44/3

=> 4(2n-1)=44 =>2n=12 => n=6

                   Now 6Cr=15 => 6Cr=6C2   or 6C4 => r=2, 4

so the problem is turned out to be 6 choose 2 or 6 choose 4

Example:  If  nCr-1= 36 nCr=84 and nCr+1=126 , then what would be the value of r ?

 Solution : Here nCr-1 / nCr =36/84 and nCr /nCr+1 =84/126 .

(n)!/{(n-r+1)!x(r-1)!} X {(r)!x(n-r)!}/(n)!=36/84

r/(n-r+1)=3/7 => 7r=3n-3r+3

=> 3n-10r=-3, and similarly from second ration we get

4n-10r=6

On solving, we get n=9, r=3

so the problem turned out to be 9 choose 3 , 9 choose 2 and 9 choose 4.

Example: Everyone in the room shakes hands with everyone. The total count of handshaking are 66. Find the number of person in the room.

nC2 =66 => n!/{2!(n-2)!}=66 => n(n-1)=132 => n=12

Solution: so the value of n is 12 implies the total number of people in the room is 12 and the problem is 12 choose 2.

Example: In a football tournament, 153 games were played. All teams played one game. find the number of groups involved in tournament.

Solution:

here nC2 =153 => n!/{2!(n-2)} = 153 => n(n-1)/2=153 => n=18

so the total number of teams participated in the tournament were 18 and the combination is 18 choose 2 .

Example During the Deepawali ceremony each club member sends greeting cards to others. If there are 20 members in the club, what would be the total number of ways greeting cards exchanged by the members.

Solution: Since two members can exchange cards each other in two ways so there is 20 choose 2 two times

2 x 20C2 =2 x (20!)/{2!(20-2)!}=2*190=380, there would be 380 ways to exchange greeting cards.

Example: Six plus ‘+’ and four minus ‘-’ symbols should be arranged in such a straight line so that no two ‘-’ symbols meet, find  the total number of ways .

 Solution: The ordering can be make as -+-+-+-+-+-+- the (-)  signs can be put in 7 vacant (pointed) place.

Hence required number of ways = 7C4 =35 .

Example: If nC21 =nC6 , then  find nC15 =?

Solution: Given nC21 =nC6

21+6=n => n=27

Hence 27C15 =27!/{15!(27-15)!} =17383860

which is the 27 choose 15.

Conclusion

Some examples are taken depending on the relations and results, as number of examples we can take on each of the result but the important thing here I want to show was how we can use any result depending on the situation if you require further reading you can go through the content or if any personal help then you can free to contact us some of the related content you can find from:

For more topics on Mathematics, please check this link.

SCHAUM’S OUTLINE OF Theory and Problems of DISCRETE MATHEMATICS

https://en.wikipedia.org/wiki/Permutation

https://en.wikipedia.org/wiki/Combination

About DR. MOHAMMED MAZHAR UL HAQUE

I am DR. Mohammed Mazhar Ul Haque , Assistant professor in Mathematics. Having 12 years of experience in teaching. Having vast knowledge in Pure Mathematics , precisely on Algebra. Having the immense ability of problem designing and solving. Capable of Motivating candidates to enhance their performance.
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