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# Notes | EduRev

## JEE : Notes | EduRev

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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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