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?
A2400 B2700 C3000 D3200
Answer: B
Explanation:
A candidate has to select $2$, $2$ and $3$ from each group, Thus there will be three possibilities
$\begin{array}{*{20}{|c|c|c|c|}} \hline
{}&{X}&{Y}&{Z} \\
& (5) & (6) & (4)\\ \hline
{Option\;1}&{2}&{2}&{3} \\[4px] \hline
{Option\;2}&{2}&{3}&{2} \\[4px] \hline
{Option\;3}&{3}&{2}&{2} \\[4px]\hline
\end{array}$
$ = \left( {{}^5{C_2} \times {}^6{C_2} \times {}^4{C_3}} \right)$ $ + \left( {{}^5{C_2} \times {}^6{C_3} \times {}^4{C_2}} \right)$ $ + \left( {{}^5{C_3} \times {}^6{C_2} \times {}^4{C_2}} \right)$
$ = (10 \times 15 \times 4)$ $+ (10 \times 20 \times 6)$ $+ (10 \times 15 \times 6)$
$= 600 + 1200 + 900$
$= 2700$
32
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?
A50 B52 C56 D60
Answer: C
Explanation:
As all green balls are similar, number of ways of slecting any number of green balls is 1. So we can omit selecting of green balls from calculation.
Three balls can be chosen in number of ways
$\begin{array}{*{20}{|c|c|c|c|}} \hline
{Option }&{White}\;(3)&{Blue}\;(4)&{Green}\;(5) \\
\downarrow & Diff. & Diff. & Similar\\ \hline
{1}&{1}&{1}&{1} \\[4px] \hline
{2}&{2}&{1}&{0} \\[4px] \hline
{3}&{1}&{2}&{0} \\[4px]\hline
{4}&{0}&{3}&{0} \\[4px]\hline
{5}&{0}&{1}&{2} \\[4px]\hline
{6}&{0}&{2}&{1} \\[4px]\hline
\end{array}$ I. 1 Blue, 1 White & 1 Green
$ = {}^4{C_1} \times {}^3{C_1} = 12$
II. 2 Blue, 1 White
= ${}^4{C_2} \times {}^3 {C_1} $ $= 18$
III. 1 Blue, 2 White
= ${}^4{C_1} \times {}^3 {C_2} $ $= 12$
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?
Explanation:
At least 1 book out of 3 Mathematics books can be selected in ${{2}}^{{3}} {{ - 1}}$ (or 7) ways.
At least 1 book out of 4 Science books can be selected in ${{2}}^{{4}} {{ - 1}}$ (or 15) ways
At least 1 book out of 5 literature books can be selected ${{2}}^{{5}} {{ - 1}}$ in (or 31) ways.
So, the total number of ways in which we can select at least one book of each type is $7 \times 15 \times 31$
34
What is the number ways of selecting 4 articles out of 10 articles? Given that 3 out of these 10 articles are identical.
A98 B90 C120 D24
Answer: A
Explanation:
Here 3 articles are identical; Hence we can omit selecting them.
$ = {}^7{C_4} + {}^7{C_3} + {}^7{C_2} + {}^7{C_1}$
$= 35 + 35 + 21 + 7$
$= 98$ ways
35
In how many ways can a pack of cards be equally divided among two players?
Explanation:
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 ${}^{52}{C_{26}}$ in ways or $\dfrac{{52!}}{{26! \times 26!}}$.
For other player we have to select 26 cards out of the remaining 26 cards, which we can do only in 1 way.
So the total number of ways to give them 26 cards each is $\dfrac{{52!}}{{26! \times 26!}}$
36
In how many ways can 10 identical items be distributed among 6 students so that each student gets at least one item?
A219 B256 C126 D${}^{15}{C_5}$
Answer: C
Explanation:
Formula, number of ways of distributing $n$ similar articles to $r$ persons so that each gets atleast one article $ = {}^{n - 1}{C_{r - 1}}$
$ = {}^{10 - 1}{C_{6 - 1}}$ $= {}^9{C_5}$ $= 126$ ways.
37
In how many ways can 10 identical items be distributed among 6 students so that each student gets any number of items?
Explanation:
Formula, number of ways of distributing $n$ similar articles to $r$ persons so that each gets atleast one article $ = {}^{n + r - 1}{C_{r - 1}}$
$ = ^{10 + 6 - 1} C_{6 - 1} $ $= {}^{15}{C_5}$
