PERMUTATION COMBINATION DEFINE 11TH CLASS
Consider the earlier example from permutation with 2 chairs were filled from a group of 4 person. we make a little change in the problem we want to select a group of 82 people and not consider the order so the arrangements a b and b a corresponding to the group. Similarly b c and c B are given in the same group the list is given as follows
AB BC CA AD BD CD
BA CB AC DA DB DC
Thus there are altogether
\[\frac{\text{^4P_2}}{\text{2}}=6\]
different groups selected. This is called the combination number of selecting a group selected. This is called the combination number of selecting a group of 2 from 4 persons denoted \[^4C_2.\]
Definition Combination:
A combination of a set of n distinct objects taken r at a time without repetion is an r-element subset of the n objects.
Note: The order of arrangement of the elements is immaterial in a combination.
if a want to choose a team of 3 players from a set of 8 different players, we first get the number ^8P_3 i.e, different ordered sets of 3 players and since any set of 3 players and since any set of 3! ordered sets, we divided \[^3P_3\] by 3!. Thus the number of combination of 3 players is
\[=\frac{\text{^8P_3}}{\text{3!}}\]
\[=\frac{\text{8!}}{\text{(8-3)!3!}}\]
COMBINATION
From n different objects of ways selecting a group or a set of r objects (without considering order) is denoted by ^nC_r or C(n,r) or nCr. It is the number of combinations of r nCr objects from n distinct objects
Theorem: \[^nC_r=\frac{\text{n!}}{\text{(n-r)!r!}}\]
Properties of Combination.
1.Consider, \[^nC_{n-r}=\frac{\text{n!}}{\text{(n-r)![n-(n-r)]!}}\]
\[=\frac{\text{n!}}{\text{(n-r)!r!}}\]
\[=^nC_r.\]
thus \[^nC_{n-r}=^nC_r , 0<= r <= n.\]
2. Consider, \[^nC_0= \frac{\text{n!}}{\text{(n-0)!)0!}}\]
\[=^nC_0\]
\[=\frac{\text{n!}}{\text{n!}}=1\]
as has been stated earlier
3. If \[^nC_r = ^nC_3\] , then either \[s=r or s=n-r\]
4. \[^nC_r=\frac{\text{nPr}}{\text{r!}}\]
5. \[^nC_r + ^nC_{r-1}=^{n+1}C_r\]
6. \[^nC_0+^nC_1+...+^nC_n=2^n\]
7. \[^nC_0+^nC_2+^nC_4+....=^nC_1+^nC_3+^nC_5+....=2^{n-1}\]
8. \[^nC_r=\left(\frac{\text{n}}{\text{r}}\right)\]
\[ ^{n-1}C_{r-1}=\left(\frac{\text{n-1}}{\text{r-1}}\right)\]
\[^{n-2}C_{r-2}=...\]
\[^nC_r\] has maximum value if
(a) \[r=\frac{\text{n}}{\text{2}}\] when n is even
(b)\[r=\frac{\text{(n-1)}}{\text{2}}\] or \[r=\frac{\text{n+1}}{\text{2}}\] when n is odd
EXAMPLES ON COMBINATION
Find the value of :
\[^nC_r=\frac{n!}{r!(n-r)!}\]
(i) \[^7C_3=\frac{7!}{3!(7-3)!}\]
\[=\frac{7\times6\times5\times4!}{3!4!}=\frac{7\times6\times5}{3!} =35\]
(ii) \[^10C_7=\frac{10!}{7!(10-7)!}\]
\[=\frac{10\times9\times8\times7!}{3!7!}\]
\[=\frac{10\times9\times8}{3\times2}=120\]
Find n and r if \[^nC_{r-1}:^nC_r:^nC_{r-1}=14:8:3\]
solution: `\frac{^nC_{r-1}}{^nC_r}`
`=\frac{14}{8}=\frac{7}{8}`
`\therefore \frac{r}{n-r+1}`
`=\frac{7}{4} and \frac{r+1}{n-r}`
`\therefore (n-r)=\left(\frac{4}{7}r\right)-1 and (n-r)`
`=\frac{3}{8}(r+1)`
`\therefore \left(\frac{4}{7}r\right)-1`
`=\frac{3}{8}(r+1)^nC_r`
`=\frac{n!}{r!(n-r)!}`
(i) \[^7C_3=\frac{7!}{3!(7-3)!}\]
\[=\frac{7\times6\times5\times4!}{3!4!}\]
\[=\frac{7\times6\times5}{3!}=35\]
(ii) \[^10C_7=\frac{10!}{7!(10-7)!}\]
\[=\frac{10\times9\times8\times7!}{3!7!}\]
\[=\frac{10\times9\times8}{3\times2}=120\]
Find n and r
if \[^nC_{r-1}:^nC_r:^nC_{r-1}=14:8:3\]
solution: `\frac{^nC_{r-1}}{^nC_r}`
`=\frac{14}{8}`
`=\frac{7}{8}`
`\therefore \frac{r}{n-r+1}`
`=\frac{7}{4} and \frac{r+1}{n-r}`
`\therefore (n-r)`
`=\left(\frac{4}{7}r\right)-1` and `(n-r)`
`=\frac{3}{8}(r+1)`
`\therefore \left(\frac{4}{7}r\right)-1`
`=\frac{3}{8}(r+1)`
\[^nC_r=\frac{n!}{r!(n-r)!}\]
(i) \[^7C_3=\frac{7!}{3!(7-3)!}\]
`=\frac{7\times6\times5\times4!}{3!4!}`
`=\frac{7\times6\times5}{3!}=35`
(ii) \[^10C_7=\frac{10!}{7!(10-7)!}\]
`=\frac{10\times9\times8\times7!}{3!7!}`
`=\frac{10\times9\times8}{3\times2}=120`
Find n and r
if \[^nC_{r-1}:^nC_r:^nC_{r-1}=14:8:3\]
solution: `\frac{^nC_{r-1}}{^nC_r}=\frac{14}{8}=\frac{7}{8}`
`\therefore \frac{r}{n-r+1}` `=\frac{7}{4}` and `\frac{r+1}{n-r}`
`\therefore (n-r)` `=\left(\frac{4}{7}r\right)-1` and `(n-r)` `=\frac{3}{8}(r+1)`
`\therefore \left(\frac{4}{7}r\right)-1` `=\frac{3}{8}(r+1)`
Combination
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| PERMUTATION COMBINATION DEFINE 11TH CLASS |
