(i) Fundamental principle of multiplication: If there are two tasks such that one of them can be completed in x way and the second task can be completed in y way, then the two tasks in succession can be completed in x×y ways. (ii) Fundamental principle of addition: If there are two tasks such that they can be performed independently in x and y ways, respectively, then either of the two tasks can be performed in x+y ways.
The possible different arrangements which can be made by taking some or all the given things are called permutation. The number of permutations of m different things taken r at a time is mPr, where mPr=mm-1m-2…m-r+1. (i) In permutation, the order of objects plays an important role. (ii) The number of all permutations of n distinct things taken all at a time is n! (iii) When all objects are distinct - (a) The number of permutations of m different objects taken r at a time when a particular object is to be always included in each arrangement is r! m-1Cr-1. (b) The number of permutations of m different things taken all at a time when n specified things always come together is n!m-n+1! (c) The number of permutations of m different things taken all at a time when n specified things never come together is m!-n!m-n+1! (iv) Permutations of objects when not all are distinct: The number of permutation of n things taken all at a time, p are alike of one kind, q are alike of second kind and r are alike of a third kind and the rest n-p+q+r are all different is n!p!q!r!
Each of the selection which can be made by some or all the given things without reference to the order of the things in each group is called a combination. The number of all combinations of n objects taken r at a time is generally denoted by Cn,r or nCr. (i) In combination, only selection is made; whereas in permutation, not only a selection is made but also an arrangement in a definite order is considered. (ii) In combination, the ordering of the selected objects is immaterial; whereas in permutation, the ordering is important. (iii) nCr= nCn-r (iv) nCr+ nCr-1= n+1Cr (v) nCx= nCy⇒x=y or x+y=n (vi) If n is even, the greatest value of nCr is nCn/2 (vii) If n is odd, the greatest value of nCr is nCn+1/2 or nCn-1/2 (viii) nCo+ nC1+…+ nCn=2n (ix) nCn+ n+1Cn+ n+2Cn+…+ 2n-1Cn= 2nCn+1 (x) nCr=nr n-1Cr-1 (xi) nCr=r+1n+1 n+1Cr+1 (xii) The number of combinations of n different things taking some or all (at least one) at a time = nC1+ nC2+ nC3+… nCn=2n-1 (xiii) The total number of selections of some or all out of p+q+r items, where p are alike of one kind, q are alike of second kind and rest r are alike of third kind is p+1q+1r+1-1. (xiv) The total number of ways of selections of one or more items from p identical items of one kind, q identical as second kind, r identical items of third kind and n different items is p+1q+1r+12n-1. (xv) The number of combinations of m different things taking r at a time: (a) When q particular things are always to be included =m-qCr-q (b) When q particular things are always to be excluded =m-qCr (c) When q particular things are always included and p particular things are always excluded =n-q-pCr-q.
Let N= P1α1. P2α2. P3α3.......... Pkαk, where P1,P2,P3,….Pk are different prime numbers and α1,α2,α3,…αk are natural numbers, then: (i) The total number of divisors of N including 1 and N is α1+1α2+1α3+1…αk+1 (ii) The total number of divisors of N excluding 1 and N is α1+1α2+1α3+1…αk+1-2 (iii) The total number of divisors of N excluding 1 or N is α1+1α2+1α3+1…αk+1-1 (iv) The sum of these divisors is P10+P11+P12+…+P1α1 P20+P21+P22+…+P2α2…Pk0+Pk1+Pk2+…+Pkαk
If n things are arranged in a row, the number of ways in which they can be deranged so that none of them occupies its original place is n!1-11!+12!-13!+......+-1n1n!
(i) The number of circular permutations of n different things taking all together is n-1! when clockwise and anti-clockwise orders are treated as different. (ii) The number of circular permutations of n different things taking all together is 12n-1! when above two orders are treated as same.
(i) The number of ways in which n distinct objects can be split into three groups containing respectively r, s and t distinct objects and r+s+t=n, is given by nCr n-rCs n-r-sCt=n!r!s!t! (ii) If we have to divide n distinct objects into l groups containing p objects each and m groups containing q objects each, then ways of group formation will be n!p!lq!ml!m! and permutation of these groups is equal to n!l+m!p!lq!ml!m!. Here, lp+mq=n.
The continued product of first n natural numbers is referred as n factorial and is represented by n!. (i) n!=nn-1n-2….3·2·1 (ii) n!=nn-1!=nn-1n-2! =nn-1n-2n-3! (iii) nn-1…n-r+1=n!n-r!
To compute the number of union of sets, firstly, sum the number of these sets separately, and then subtract the number of all pairwise intersections of the sets, then add back the number of the intersections of triples of the sets, and so on, up to the intersection of all sets. If A and B are not disjoint, then we get the simplest form of the Inclusion-Exclusion principle, i.e. nA∪B=nA+nB-nA∩B
(ii) Number of total triangles formed by joining n points in a plane of which m are collinear is nC3-mC3. (iii) Number of total diagonals in a polygon of n sides is nC2-n
(i) If there are l objects of one kind, m objects of second kind, n objects of third kind and so on; then the number of ways of choosing r objects out of these (i.e., l+m+n+.....) objects is the coefficient of xr in the expansion of 1+x+x2+x3+…xl1+x+x2+x3+…xm1+x+x2+x3+…xn….. Further, if one object of each kind is to be included, then the number of ways of choosing r objects out of these objects (i.e., l + m + n +....) is the coefficient of xr in the expansion of x+x2+x3+…xlx+x2+x3+…xmx+x2+x3+…xn….. (ii) If there are l objects of one kind, m objects of second kind, n objects of third kind and so on; then the number of possible arrangements / permutations of r objects out of these objects (i.e., l + m + n +.....) is the coefficient of xr in the expansion of r!1+x1!+x22!+......+xll!1+x1!+x22!+......+xmm!
(i) The number of non-negative integral solutions for the equation x1+x2+x3 ....xr = n is n+r-1Cr-1. Here, the value of any variable can be zero. (ii) The number of positive integral solutions for the equation x1+x2+x3 ....xr = n is n-1Cr-1. Here, the minimum value for any variable is 1.
Let X and Y be two finite sets having m and n elements respectively. Then, each element of set X can be associated to any one of n elements of sets Y. So, total number of functions from set X to set Y is nm. Number of one-one (injection) functions: If A and B are finite sets having m and n elements respectively, then the number of one-one functions from A to B=nPm if n≥m, otherwise 0. Number of onto (surjection) functions: If A and B are two sets having m and n elements, respectively, such that 1≤n≤m, then the number of onto functions from A to B is Σr=0n-1r nCr(n-r)m, otherwise 0. Indian Navy SSR Agniveer Admit Card Released! The Indian Navy had released the official notification for the Indian Navy SSR Agniveer Exam 2022. A total of 2800 vacancies are released for the recruitment process of Indian Navy SSR AA. The application process has started on 1st July and candidates can apply till 27th July 2022. The selection process includes a Written Test, Physical Fitness Test (PFT), and Medical Examination. With a salary of Rs. 30,000, it is a golden opportunity for job seekers.
I doubt whether you have worded your question correctly as according to your question the answer should come out as 576 and not 288. Anyway an approach that could lead to 288 is as follows: Let's number girls and boys as : G1, G2, G3, G4 and B1, B2, B3 respectively. Since a girl has to sit next to a girl only therefore all the girls would sit together G1,G2,G3,G4 or G2,G3,G1,G4 etc. Now these girls can be arranged in 4! = 24 ways. Now we have 3 boys who can be made to sit together in 3! = 6 ways. Finally we have Girls (G) and Boys (B) who have to be seated together and they can be seated in 2! = 2 ways (consider 4 girls as one group = G and 3 boys as one group = B) From above three steps we get the total number of ways as : 4! * 3! * 2! = 288 |