Permutations and Combinations is one of the most important chapters of Algebra in the JEE syllabus and other engineering exams. For JEE Mains, it has 4% weightage and for JEE Advanced, it has 5% weightage..

**Q1.**The number of ways in which we can get a score of 11 by throwing three dice is

Solution

Dice is marked with numbers 1, 2, 3, 4, 5, 6. If the sum of dice in three throws is 11, then observations must be 1, 4, 6;…1, 5, 5;…2, 3, 6;…2, 4, 5;… 3, 3, 5; …3, 4, 4

We can get this observation in 3! + 3!/2! + 3! + 3! + 3!/2! + 3!/2! = 27 ways

Dice is marked with numbers 1, 2, 3, 4, 5, 6. If the sum of dice in three throws is 11, then observations must be 1, 4, 6;…1, 5, 5;…2, 3, 6;…2, 4, 5;… 3, 3, 5; …3, 4, 4

We can get this observation in 3! + 3!/2! + 3! + 3! + 3!/2! + 3!/2! = 27 ways

**Q2.**There are two bags each containing balls. If a man has to select equals number of balls from both the bags the number of ways in which he can do so if he must chose at least one ball from each bag is

Solution

The number of ways of selecting r(0 ≤ r ≤ m) balls out of m is

Therefore, the number of ways if selecting balls from each of the bag is (

(

[∵ (

The number of ways of selecting r(0 ≤ r ≤ m) balls out of m is

^{m}C_{r}.Therefore, the number of ways if selecting balls from each of the bag is (

^{m}C_{r})^{2}, Further the number of ways of selecting equal number of balls from each of the two bags, choosing at least one from each bag, is(

^{m}C_{1})^{2}+ (^{m}C_{2})^{2}+ ... + (^{m}C_{m})^{2}=^{2m}C_{m}- 1[∵ (

^{m}C_{0})^{2}+ (^{m}C_{1})^{2}+ ... + (^{m}C_{m})^{2}=^{2m}C_{m}]**Q3.**Eight chairs are numbered 1 to 8. Two women and three men wish to occupy one chair each. First, the women choose the chairs from amongst the chairs marked 1 to 4, and then the men select the chairs from amongst the remaining. The number of possible arrangements is

Solution

Two women can choose two chairs out of 1, 2, 3, 4 in

Therefore, total number of possible arrangements is

Two women can choose two chairs out of 1, 2, 3, 4 in

^{4}C_{2}ways, and can arrange among themselves in 2! ways. Three men can choose 3 chairs out of 6 remaining chairs in^{6}C_{3}ways and can arrange themselves in 3! ways.Therefore, total number of possible arrangements is

^{4}C_{2}X 2! X^{6}C_{3}X 3! =^{4}P_{2}X^{6}P_{3}

**Q4.**n is selected from the set {1,2,3,...,10} and the number 2

^{n}+ 3

^{n}+ 5

^{n}is formed. Total number of ways of selecting so that the formed number is divisible by 4 is equal to

**Q5.**Let A = {x

_{1}, x

_{2}, x

_{3}, ..., x

_{7}}, B = {y

_{1}, y

_{2}, y

_{3}}. The total number of functions f:A → B that are on to and there are exactly three element in such that f(x) = y

_{2}is equal to

Solution

Three elements from set ‘A’ can be selected in

Three elements from set ‘A’ can be selected in

^{7}C_{3}ways. Their image has to be Remaining 2 images can be y_{2}assigned to remaining 4 pre-images in 2^{4}ways. But the function is onto, hence the number of ways is 2^{4}- 2. Then the total number of functions is^{7}C_{3}X 14 = 490**Q6.**Number of ways in which 25 identical things be distributed among five persons if each gets odd number of things is

**Q7.**The number of integral solutions of x + y + z = 0 with x ≥-5, y≥-5,z≥-5, is

Solution

Let x = p, y = q - 5 and z = r - 5, where p,q,r ≥ 0

Then the given equation reduces to p + q + r = 15 -- (1)

Now, we have to find non-negative integral solution of Eq. (1). The total number of such solutions is

Let x = p, y = q - 5 and z = r - 5, where p,q,r ≥ 0

Then the given equation reduces to p + q + r = 15 -- (1)

Now, we have to find non-negative integral solution of Eq. (1). The total number of such solutions is

^{15+3-1}C_{3-1}=^{17}C_{2}= 136**Q8.**The total number of five-digit numbers of different digits in which the digit in the middle is the largest is

**Q9.**The total number of ways in which persons can be divided into couples is

Solution

Here, we are dividing 2n people in group of 2 each, and we are concerned with mere grouping. Hence, the required number of ways is 2n!/(n!(2!)

Here, we are dividing 2n people in group of 2 each, and we are concerned with mere grouping. Hence, the required number of ways is 2n!/(n!(2!)

^{n})**Q10.**In how many different ways can the first 12 natural numbers be divided into three different groups such that numbers in each group are in A.P.?

Solution

No group of four members from the first 12 natural number can have the common difference 4.

If one group including 1 is selected with the common difference 1, then the other two group can have the common difference 1 or 2.

If one group including 1 is selected with the common difference 2, then one of the other two groups can have the common difference 2 and the remaining group will have common difference 1.

If one group including 1 is selected with common difference 3, then the other two group can have the common difference 3 Therefore, the required number of ways is 2 + 1 + 1 = 4

No group of four members from the first 12 natural number can have the common difference 4.

If one group including 1 is selected with the common difference 1, then the other two group can have the common difference 1 or 2.

If one group including 1 is selected with the common difference 2, then one of the other two groups can have the common difference 2 and the remaining group will have common difference 1.

If one group including 1 is selected with common difference 3, then the other two group can have the common difference 3 Therefore, the required number of ways is 2 + 1 + 1 = 4