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JEE Revision Notes

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  • If certain objects are to be arranged in such a way that the order of objects is not important, then the concept of combinations is used.
  • The number of combinations of n things taken r (0 < r < n) at a time is given by nCr= n!/r!(n-r)!
  • The relationship between combinations and permutations is nCr = nPr/r!
  • The number of ways of selecting r objects from n different objects subject to certain condition like:
    1. k particular objects are always included =  n-kCr-k
    2. k particular objects are never included =  n-kCr
  • The number of arrangement of n distinct objects taken r at a time so that k particular objects are
    1. Always included = n-kCr-k.r!,
    2. Never included = n-kCr.r!.
  • In order to compute the combination of n distinct items taken r at a time wherein, the chances of occurrence of any item are not fixed and may be one, twice, thrice, …. up to r times is given by n+r-1Cr
  • If there are m men and n women (m > n) and they have to be seated or accommodated in a row in such a way that no two women sit together then total no. of such arrangements
    = m+1Cn. m! This is also termed as the Gap Method.
  • If there is a problem that requires n number of persons to be accommodated in such a way that a fixed number say ‘p’ are always together, then that particular set of p persons should be treated as one person. Hence, the total number of people in such a case becomes (n-m+1). Therefore, the total number of possible arrangements is (n-m+1)! m! This is also termed as the String Method.  
  • Let there be n types of objects with each type containing at least r objects. Then the number of ways of arranging r objects in a row is nr. 
  • The number of selections from n different objects, taking at least one
    = nC1 + nC2 + nC+ ... + nCn = 2n - 1.
  • Total number of selections of zero or more objects from n identical objects is n+1.
  • Selection when both identical and distinct objects are present: 
  • The number of selections, taking at least one out of a1 + a2 + a3 + ... an + k objects, where a1 are alike (of one kind), a2 are alike (of second kind) and so on ... an are alike (of nth kind), and k are distinct
    = {[(a1 + 1)(a2 + 1)(a3 + 1) ... (an + 1)]2k} - 1.
  • Combination of n different things taken some or all of n things at a time is given by 2– 1. 
  • Combination of n things taken some or all at a time when p of the things are alike of one kind, q of the things are alike and of another kind and r of the things are alike of a third kind
    = [(p + 1) (q + 1)(r + 1)….] – 1
  • Combination of selecting s1 things from a set of n1 objects and sthings from a set of nobjects where combination of s1 things and s2 things are independent is given by n1Cs1 x n2Cs2  
  • Some results related to nC
    1. nC= nCn-r
    2. If nC= nCk, then r = k or n-r = k
    3. nC+ nCr-1 = n+1Cr
    4. nCr = n/r  n-1Cr-1
    5. nCr/nCr-1 = (n-r+1)/ r
    6. If n is even nCr is greatest for r = n/2
    7. If n is odd, is greatest for r = (n-1) /2, (n+1)/2
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