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ʜʟᴏ ɢᴜyꜱ ᴀᴀᴩ ꜱᴀʙʜɪ ʙᴀʀ ʙᴀʀ ᴩᴜᴄʜ ʀᴀʜᴇ ᴛʜᴇ ᴋɪ ɴᴇᴇᴛ ᴋɪ ᴩᴀᴅʜᴀɪ ᴋʜᴀ ꜱᴇ ᴋᴀʀᴇ ᴛᴏʜ ᴍᴇɴᴇ ᴇxᴀᴍɪɴᴇ ᴋɪyᴀ ʜᴀɪ ᴀᴜʀ yᴇ ᴀᴩᴩ ᴀᴄᴄʜᴀ ʟᴀɢᴀ ʜᴀɪ yᴇᴀʜ ꜱᴀʀᴇ ʟᴇᴄᴛᴜʀᴇ ꜰʀᴇᴇ ʜᴀi ᴏʀ ʟɪᴠᴇ ᴄʟᴀꜱꜱ ᴄʜᴀʟᴛᴇ ʜᴀɪ,
ᴛᴇꜱᴛ ꜱᴇʀɪᴇꜱ ᴏʀ ᴍᴏᴄᴋ ᴛᴇꜱᴛ ʙʜɪ ʜᴀɪ ᴊᴀʟᴅɪ ᴅᴏᴡɴʟᴏᴀᴅ ᴋᴀʀᴏ ᴀᴜʀ ꜱʜᴀʀᴇ ᴋᴀʀᴏ ɪꜱꜱ ᴀᴩᴩ ᴋᴏ ʙᴀʜᴜᴛ ᴀᴄᴄʜᴀ ᴀᴩᴩ ʜᴀɪ yᴇ
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ᴊᴀᴏ ɢᴜyꜱ ꜰᴀꜱᴛ ᴅᴏᴡɴʟᴏᴀᴅ ᴛʜɪꜱ ᴀᴩᴩ
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⭕️Algebra - Revision Notes on Permutations⭕️
➖The concept of permutation is used for the arrangement of objects in a specific order i.e. whenever the order is important, permutation is used.
➖The total number of permutations on a set of n objects is given by n! and is denoted as nPn = n!
➖The total number of permutations on a set of n objects taken r at a time is given by nPr = n!/ (n-r)!
➖The number of ways of arranging n objects of which r are the same is given by n!/ r!
➖If we wish to arrange a total of n objects, out of which ‘p’ are of one type, q of second type are alike, and r of a third kind are same, then such a computation is done as n!/p!q!r!
➖Al most all permutation questions involve putting things in order from a line where the order matters. For example ABC is a different permutation to ACB.
➖The number of permutations of n distinct objects when a particular object is not to be considered in the arrangement is given by n-1Pr
➖The number of permutations of n distinct objects when a specific object is to be always included in the arrangement is given by r.n-1Pr-1.
➖If we need to compute the number of permutations of n different objects, out of which r have to be selected and each object has the probability of occurring once, twice or thrice… up to r times in any arrangement is given by (n)r.
➖Circular permutation is used when some arrangement is to be made in the form of a ring or circle.
➖When ‘n’ different or unlike objects are to be arranged in a ring in such a way that the clockwise and anticlockwise arrangements are different, then the number of such arrangements is given by (n – 1)!
➖If n persons are to be seated around a round table in such a way that no person has similar neighbor then it is given as ½ (n – 1)!
➖The number of necklaces formed with n beads of different colors = ½ (n – 1)!
➖nP0 =1
➖nP1 = n
➖nPn = n!/(n-n)! = n! /0! = n! /1= n!
