What is the number of ways in which the word able can be arranged so that the vowels occupy even places?

Text Solution

Solution : The even places can be occupied only by consonants (RTCL) we select `3` out of `4` to occupy these `3` places = `4_(C_3)`=4 They can themselves arrange in `3!` ways =`6` ways total =`6.4` ways =`24` ways For the odd places remaining `4` letters will arrange themselves in 4! ways =`24` ways So, total =`24.24` ways =`576` ways

1. Fundamental Principle of Counting:

(i) Fundamental principle of multiplication:

If there are two tasks such that one of them can be completed in x ways and the second task can be completed in y ways, 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.

2. Permutation:

The possible different arrangements which can be made by taking some or all of given things is 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!

3. Combinations:

Each of the selection which can be made by some or all of 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.
Then nCr=n!r!n-r!0≤r≤n= nPrr!

(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, then the greatest value of nCr is nCn/2

(vii) If n is odd, then 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

4. Number of Divisors:

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

5. Derangements:

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!

6. Circular Permutation:

(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.

7. Division and Distribution of Objects:

(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!)l(q!)ml!m! and permutation of these groups is equal to n!(l+m)!(p!)l(q!)ml!m!. Here, lp+mq=n.

8. Factorial Notation:

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!

9. Principle of Inclusion and Exclusion:

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:

10. Geometrical Application of Combinations:

(i) Number of total different straight lines formed by joining n points in a plane of which m are collinear is nC2- mC2+1

(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

11. Multinomial Theorem:

(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 object 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!

12. Number of Integral Solutions:

(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.

13. Number of Functions:

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 ∑r=0n-1r nCr(n-r)m, otherwise 0.