Free Online Preparation for CAT with Minglebox e-CAT Prep. Cover basic concepts of Permutations and Combinations under Quantitative Aptitude for MBA Entrance Exam Preparation with Study material, solved examples and tests prepared by CAT coaching experts.
The number of arrangements of 'n' distinct object into 'n' distinct position is given by n,n-1,n-2,........3,2,1, which is often denoted as n! ( read as n 'factorial'). The number of arrangements is also know as “Permutations". Thus number of permutation of n distinct object into n distinct places is denoted as , read as permutations of n distinct objects taken n at a time.
The number of Permutations of n distinct objects taken from a group of 'r' distinct objects, in n distinct place is given by =
The number of Permutations of n object out of which m are of one type, q are of second type etc. Such that n ≠ m ≠q …. is given by
The number of Permutation of n distinct objects taken all as times around a circle is given by (n-1)! And accordingly as clockwise and anticlockwise arrangements are treated as different or same. This formula is valid for a necklace.
Number of arrangements of n objects, such that any p out of those occur together is given by
(n-p+1) ! * p !.
Number of arrangements of n distinct things, when any object can be repeated any number of times is
given by .
Number of combinations of r distinct things taken out of n distinct things is given by
Please note in case of combinations also known as selections the order in which things are chosen is not important. Thus if we consider 4 distinct alphabets viz. A,B,C & D then no of permutations of any 3 alphabets out of is given by ABC
ADC, etc whereas no. of selections is ABC,ABD,ACD
and BCD i.e. 4 only.
Number of ways of selecting r things out of n distinct things is same as number of ways of selecting n-r things out of n things. i.e.
No of ways in which none or some objects can be chosen out of n distinct of objects is given by
The total number of ways in which n identical things can be distributed into r distinct boxes, such that one or more than one box may remain empty, but not all the boxes can be empty is given by
If blank boxes are not allowed i.e each box has at least one object then no of ways is given by
Number of non-negative solutions of is given by
Number of Non – negative solution of x 1 +x 2 …......... x r = n is given by
A 3 digit number is formed using the digits 2,3 and 4 without repeating any one of them what is the sum of all such possible numbers ?
Let us first find the total number of 3 digit numbers which be formed using the digits 2,3 and 4. The total number of such numbers is 3 ! = 6 now out of these 6 numbers we notice that number 2 appears in the hundreds place as well as in tens place and units place. So let us see how many numbers are there where 2 appears in the hundreds place we find there are exactly 2! numbers. Similarly numbers 3 and 4 appear 2 ! times in hundreds place. Hence if split each of the 6 numbers as say for example 234 as
then if we add all the numbers we see that numbers 2,3,4appear in hundreds place
2! times each. Sum of hundreds digits of all the numbers – Similarly sum of all the
numbers in the tens and units digit is given by and
therefore sum of all the numbers is
Atul has 9 friends ; 4 males and 5 females. In how many ways can he invite them, if he wants to have exactly 3 females in the invites ?
The 3 girls which are to be invited can be selected in ways. Also the remaining 4 males none, all
or some can be invited in ways. Hence Total no of ways Atul can invite his friends is
How many numbers can be formed from 2,3,4,5,6 (without repetition), when the digit at tens place must be greater than that in the hundreds place ?
Total number of numbers which can be formed using the digits 2,3,4,5,6 only once (without repetition) is 5! = 120. Now since digits are not be repeated hence if we consider any number out the 120 so formed, then either the tens digits will be greater then hundreds digits or smaller than it. Hence we have 2 possibilities for any number. Therefore half the numbers out of 120 will satisfy this and other half will not. Hence answer is 60.
A polygon has 35 diagonals find the number of sides of the polygon.
The number of diagonals of a polygon having n sides is given by . But no of diagonals is 35
Solving we get n = 10. Hence number of sides of polygon is 10.