Problems on Selection:
1. For an examination, a candidate has to select 7 subjects from three different groups X, Y and Z which contain 5, 6 and 4 subjects respectively. In how many different ways can a candidate make his selection if he has to select at least 2 subjects from each group?
Problems on distribution:
1. For an examination, a candidate has to select 7 subjects from three different groups X, Y and Z which contain 5, 6 and 4 subjects respectively. In how many different ways can a candidate make his selection if he has to select at least 2 subjects from each group?
A candidate has to select 2, 2 and 3 from each group. Thus there will be three possibilities since from any of the three groups we can select 3 subjects while we select 2 subjects each from the remaining two groups.
$^{\rm{5}} {\rm{C}}_{\rm{2}} {\rm{ \times }}^{\rm{6}} {\rm{C}}_{\rm{2}} {\rm{ \times }}^{\rm{4}} {\rm{C}}_{\rm{3}} {\rm{ + }}^{\rm{5}} {\rm{C}}_{\rm{2}} {\rm{ \times }}^{\rm{6}} {\rm{C}}_{\rm{3}} {\rm{ \times }}^{\rm{4}} {\rm{C}}_{\rm{2}} {\rm{ + }}^{\rm{5}} {\rm{C}}_{\rm{3}} {\rm{ \times }}^{\rm{6}} {\rm{C}}_{\rm{2}} {\rm{ \times }}^{\rm{4}} {\rm{C}}_{\rm{2}} $
= 10 x 15 x 4 + 10 x 20 x 6 + 10 x 15 x 6
= 600 + 1200 + 900
= 2700
2. A box contains 3 different White balls, 4 different Blue balls, 5 similar Green balls. In how many ways can 3 balls be drawn the box so that at least 1 blue ball is always included in the draw?
Three balls can be chosen in no.of ways
i. 1 Blue, 1 White & 1 Green
= $^{\rm{4}} {\rm{C}}_{\rm{1}} {\rm{ \times }}^{\rm{3}} {\rm{C}}_{\rm{1}} $ x 1 = 12
ii. 2 Blue, 1 White
= $^{\rm{4}} {\rm{C}}_{\rm{2}} {\rm{ \times }}^{\rm{3}} {\rm{C}}_{\rm{1}} $ = 18
iii. 1 Blue, 2 White
$^{\rm{4}} {\rm{C}}_{\rm{1}} {\rm{ \times }}^{\rm{3}} {\rm{C}}_{\rm{2}} $ = 12
iv. 3 Blue
$^{\rm{4}} {\rm{C}}_{\rm{3}} $ = 4
v. 1 Blue, 2 Green
= $^{\rm{4}} {\rm{C}}_{\rm{1}} {\rm{ \times 1 = 4}}$
vi. 2 Blue, 1 Green
= $^{\rm{4}} {\rm{C}}_{\rm{2}} $ x 1 = 6 x 1 = 6
Total ways = 12 + 18 + 12 + 4 + 4 + 6 = 56 ways.
3. Out of 3 mathematics books, 4 science books, and 5 literature books (all are different), how many different collections can be made by taking atleast one of each kind?
At least 1 book out of 3 Mathematics books can be selected in ${\rm{2}}^{\rm{3}} {\rm{ - 1}}$ (or 7) ways, similarly at least 1 book out of 4 Science books can be selected in ${\rm{2}}^{\rm{4}} {\rm{ - 1}}$ (or 15) ways and at least 1 book out of 5 literature books can be selected ${\rm{2}}^{\rm{5}} {\rm{ - 1}}$ in (or 31) ways. So, the total number of ways in which we can select at least one book of each type is 7x15x31
4. What is the number ways of selecting 4 articles out of 10 articles? Given that 3 out of these 10 articles are identical.
Here 3 articles are identical; hence they have no.of ways equal to
$^{\rm{7}} {\rm{C}}_{\rm{4}} {\rm{ + }}^{\rm{7}} {\rm{C}}_{\rm{3}} {\rm{ + }}^{\rm{7}} {\rm{C}}_{\rm{2}} {\rm{ + }}^{\rm{7}} {\rm{C}}_{\rm{1}} $
= 35 + 35 + 21 + 7
= 98 ways
Problems on distribution:
5. In how many ways can a pack of cards be equally divided among two players?
Here we have a peck of 52 cards and we have to distribute these cards between two players so that either of them has 26 cards. For first player we have to select 26 cards out of 52 cards and this can be done $^{{\rm{52}}} {\rm{C}}_{{\rm{26}}} $ in ways or 52!/26!26!. For other player we have to select 26 cards out of remaining 26 cards, which we can do only in 1 way. So the total number of ways to give them 26 cards each is 52!$\left( {{\rm{26!}}} \right)^{\rm{2}} $
6. In how many ways can 10 identical items be distributed among 6 students so that each student gets at least one item?
After giving one item to each of the six children, remaining items can be given in
$^{4 + 6 - 1} C_{6 - 1} $ ways = $^{\rm{9}} {\rm{C}}_{\rm{5}} $ = 126ways.
7. In how many ways a pack of cards be equally divided among (i) 4 players (ii) 4 groups?
(i) $\displaystyle\frac{{52!}}{{{{\left( {13!} \right)}^4}}}$
(ii) $\displaystyle\frac{{52!}}{{4!{{\left( {13!} \right)}^4}}}$
