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We don't always want to select some objects and arrange them in order.sometimes, we just want to select them no matter what the order. Such a way of selecting different items from a collection of items,in which their arrangement or order doesn't matter is known as Combination. Example: suppose you want to select a basketball team of 5, their order doesn't matter. It doesn't matter which player is chosen first or last in a combination.
Denoting Combination:From 'n' objects the combination of taking r at a time is given by,$$\;C(n,r)\;or\;^nC_r\;or_nC_n\;or\;C_{n,r}\;or\;\binom{n}{r}$$
Difference between permutation and Combination
In a combination, only a group is made and the order in which it is made is unimportant or not counted. In the case of permutation not only is the group formed but each arrangement of the group with the same members is also counted as a different unit. We can understand it by the following example: The combination of 3 alphabets P, Q, R taking 2 at a time is either PQ or QR or PQ.( In permutation, PQ and QP are taken as two different units but in combination PQ and QP are the same as their order doesn't matter.) Thus, C(3,2) = 3Note:.The word 'arrangements' is used for Permutation and the word "selection" is used to denote combination.
We can also find the value of combination using Factorial notation, The number of all the combinations of n objects taking r at a time is given by
$$\;C(n,r)\;or\;\binom{n}{r}\;^nC_r\;=\;\frac{n!}{(n-r)! r!}$$
Proof:To prove that,$$\;^nC_r\;=\;\frac{n!}{(n-r)! r!}$$
Let x be the total number of the combination. Since each combination has r! objects, each combination will produce r! permutations or arrangements when arranged among themselves. Therefore, x combinations will produce x.r! permutations, Which is same as the total number of permutations of n objects taken r at a time. So,
$$\therefore\;^nP_r\;=\;x.r!\;$$
$$\;So,\;x\;=\;\frac{^nP_r}{r!}\;=\;\frac{n!}{(n-1)!r!}$$
Since, x resembles the combination^{n}C_{r}.$$\;^nC_r\;=\;\frac{n!}{(n-r)! r!}$$
Statement 1:putting r=0 we get
$$\;^nC_0\;=\;\frac{n!}{(n-0)! 0!}\;=\;\frac{n!}{n!}\;=\;1$$
Statement2:putting r= 1 we get
$$\;^nC_1\;=\;\frac{n!}{(n-1)! 1!}\;=\;\frac{n!}{(n-1)!}\;=\;n$$
Statement3:putting r= n we get
$$\;^nC_n\;=\;\frac{n!}{(n-n)! n!}\;=\;\frac{n!}{0!.n!}\;=\;1$$
Statement 4:putting r=0 we get
$$\;^nC_{n-r}\;=\;\frac{n!}{(n-(n - r))! (n-r)!}\;=\;\frac{n!}{r!(n-r)!}\;=\;^nC_r$$
$$\therefore\;^nC_{n-r}\;=\;^nC_r$$
Statement 5:$$\;^nC_r\;+\;^nC_{r-1}\;=\;^{n+1}C_r$$
Proof:
$$\;^nC_r\;+\;^nC_{r-1}\;=\;\frac{n!}{(n-r)!r!}\;+\;\frac{n!}{(n-(r-1))!(r-1)!}$$
$$\;=\;\frac{n!}{(n-r)!(r-1)!}\;\left(\frac{1}{r}\;+\;\frac{1}{n-r+1}\right)$$
$$\;=\;\frac{n!}{(n-r)!(r-1)!}\;\left(\frac{n -r + 1 + r}{r(n - r +1)}\right)$$
$$\;=\;\frac{n!}{(n-r)!(r-1)!}\;\left(\frac{n + 1 }{r(n - r +1)}\right)$$
$$\;=\;\frac{n!\times(n+1)}{(n-r)!(n-r+1).r(r-1)!}\;$$
$$\;=\;\frac{(n+1)!}{(n-r+1)!r!}\;=\;^{n+1}C_r$$
Restricted Combinations:
Example:In how many ways can a basketball team of 5 be chosen out of 10 players? if
(a)A particular player is always chosen,
(b)A particular is never chosen.
Ans:
(a) A particular player is always chosen, it means that 4 players are selected out of the remaining 9 players.
Required number of ways = ^{10-1}C_{5-1} = ^{9}C_{4}
$$\;=\;\frac{9!}{5!4!}\;=\;ways$$
(b)A particular player is never chosen, it means that 5 players are selected out of 9 players.
Required number of ways = ^{9}C_{5}
$$\;=\;\frac{9!}{5!4!}\;=\;ways$$
Taken reference from
( Basic mathematics Grade XII and A foundation of Mathematics Volume II and Wikipedia.com )
Difference between permutation and Combination
In a combination, only a group is made and the order in which it is made is unimportant or not counted. In the case of permutation not only is the group formed but each arrangement of the group with the same members is also counted as a different unit. We can understand it by the following example: The combination of 3 alphabets P, Q, R taking 2 at a time is either PQ or QR or PQ.( In permutation, PQ and QP are taken as two different units but in combination PQ and QP are the same as their order doesn't matter.) Thus, C(3,2) = 3Note:.The word 'arrangements' is used for Permutation and the word "selection" is used to denote combination.
Restricted Combinations:
ASK ANY QUESTION ON Combination
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