Permutation and Combination MCQ Questions & Answers in Algebra | Maths

Given, $$^n{C_r} + {\,^n}{C_{r - 1}} + {\,^{n + 1}}{C_{r - 1}} + {\,^{n + 2}}{C_{r - 1}} + ..... + {\,^{2n}}{C_{r - 1}} = {\,^{2n + 1}}{C_{{r^2} - 132}}$$
$$\eqalign{ & \Rightarrow {\,^{n + 1}}{C_r} + {\,^{n + 1}}{C_{r - 1}} + ..... + {\,^{2n}}{C_{r - 1}} = {\,^{2n + 1}}{C_{{r^2} - 132}} \cr & .................................................. \cr & \Rightarrow {\,^{2n}}{C_r} + {\,^{2n}}{C_{r - 1}} = {\,^{2n + 1}}{C_{{r^2} - 132}} \cr & \Rightarrow {\,^{2n + 1}}{C_r} = {\,^{2n + 1}}{C_{{r^2} - 132}} \cr & \Rightarrow {r^2} - r - 132 = 0 \cr & \Rightarrow \left( {r - 12} \right)\left( {r + 11} \right) = 0 \cr & \Rightarrow r = 12 \cr & \Rightarrow n \geqslant 12 \cr} $$
So, minimum value of $$n = 12.$$

Total number of numbers without restriction $$ = {2^5}.$$
Two numbers have all the digits equal. So, the required number of numbers $$ = {2^5} - 2.$$

The numbers are made of 1, 2, 3, 4, 5 or 0, 1, 2, 4, 5.
∴ the required number of 5-digit numbers $$ = 5!\, + \left( {^5{P_5} - {\,^4}{P_4}} \right).$$

If all the letters are not in the right envelopes, then at least two letters must be in wrong envelopes.
$$\therefore x = 6! - 1 = 719.$$
Now $$y =$$  number of ways so that all the letters are in wrong envelopes
$$\eqalign{ & = 6!\left\{ {1 - \frac{1}{{1!}} + \frac{1}{{2!}} - \frac{1}{{3!}} + \frac{1}{{4!}} - \frac{1}{{5!}} + \frac{1}{{6!}}} \right\}\left[ {{\text{Deragement formula}}} \right] \cr & = 360 - 120 + 30 - 6 + 1 = 265 \cr & \therefore x - y = 454 \cr} $$

We know that a number is divisible by 3 only when the sum of the digits is divisible by 3. The given digits are $$0, 1, 2, 3, 4, 5 .$$
Here the possible number of combinations of $$5$$ digits out of $$6$$ are $$^5{C_4} = 5,$$   which are as follows -
$$1 + 2 + 3 + 4 + 5 = 15 = 3 \times 5$$
$$0 + 2 + 3 + 4 + 5 = 14$$     (not divisible by $$3$$ )
$$0 + 1 + 3 + 4 + 5 = 13$$     (not divisible by $$3$$ )
$$0 + 1 + 2 + 4 + 5 = 12 = 3 \times 4$$
$$0 + 1 + 2 + 3 + 5 = 11$$     (not divisible by $$3$$ )
$$0 + 1 + 2 + 3 + 4 = 10$$     (not divisible by $$3$$ )
Thus the number should contain the digits $$1, 2, 3, 4, 5$$   or the digits $$0, 1,2, 4, 5.$$
Taking $$1, 2, 3, 4, 5,$$   the $$5$$ digit numbers are $$= 5! = 120$$
Taking $$0, 1, 2, 4, 5,$$   the $$5$$ digit numbers are $$= 5! - 4! = 96$$
∴ Total number of numbers $$= 120 + 96 = 216$$

We have, $$^{n - 1}{C_4} - {\,^{n - 1}}{C_3} - \frac{5}{4} \cdot {\,^{n - 2}}{P_2} < 0$$
$$\eqalign{ & \Rightarrow \frac{{\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 4} \right)}}{{4!}} - \frac{{\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)}}{{3!}} - \frac{5}{4}\left( {n - 2} \right)\left( {n - 3} \right) < 0 \cr & \Rightarrow \left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 11} \right)\left( {n + 2} \right) < 0 \cr & \Rightarrow \left( {n - 2} \right)\left( {n - 3} \right)\left( {n - 11} \right) < 0\left[ {\because n + 2 > 0{\text{ for }}n \in N} \right] \cr & \Rightarrow n \in \left( { - \infty ,2} \right) \cup \left( {3,11} \right) \cr} $$

$$\eqalign{ & \Rightarrow n \in \left( {0,2} \right) \cup \left( {3,11} \right) \cr & \Rightarrow n = 1,4,5,6,7,8,9,10 \cr} $$
But $$^{n - 1}{C_4}$$  and $$^{n - 2}{P_2}$$  both are meaningful for $$n \geqslant 5.$$
Hence, $$n = 5,6,7,8,9,10.$$

The following are the number of possible choices:
$$^{52}{C_{18}} \times {\,^{35}}{C_2}$$   (18 families having atmost 2 children and 2 selected from other type of families)
$$^{52}{C_{19}} \times {\,^{35}}{C_1}$$   (19 families having atmost 2 children and 1 selected from other type of families)
$$^{52}{C_{20}}$$  (All selected 20 families having atmost 2 children). Hence, the total number of possible choices is :
$$ = {\,^{52}}{C_{18}} \times {\,^{35}}{C_2} + {\,^{52}}{C_{19}} \times {\,^{35}}{C_1}\, + {\,^{52}}{C_{20}}$$

Ways : \[\begin{array}{*{20}{c}} \times \\ 2 \end{array}\,\,\begin{array}{*{20}{c}} \times \\ 3 \end{array}\,\,\begin{array}{*{20}{c}} \times \\ 3 \end{array}\,\,\begin{array}{*{20}{c}} \times \\ 3 \end{array}\,\,\begin{array}{*{20}{c}} \times \\ 3 \end{array}\,\,\begin{array}{*{20}{c}} \times \\ 3 \end{array}\]
0 cannot go in the units place.
∴ the required number of numbers \[ = 2 \times {3^5}.\]

As shown in figure 1, 2 and $$X$$ are the three boys and 3, 4 and $$Y$$ are three girls, Boy $$X$$ will have neighbours as boys 1 and 2 and the girl $$Y$$ will have neighbours as girls 3 and 4.

1 and 2 can be arranged in $$P\left( {2,2} \right)$$  ways
$$ = 2! = 2 \times 1 = 2{\text{ ways}}$$
Also, 3 and 4 can be arranged in $$P\left( {2,2} \right)$$  ways
$$ = 2! = 2 \times 1 = 2{\text{ ways}}$$
Hence, required no. of permutations $$= 2 \times 2 = 4$$

The number of ways of distributing 10 identical balls in 4 distinct boxes such that no box empty is same as the number of ways of selecting $$(r - 1)$$  places out of $$(n - 1)$$  different places, that is $$^{n - 1}{C_{r - 1}}.$$
Hence required number of ways $$^{10 - 1}{C_{4 - 1}} = {\,^9}{C_3}$$
∴ Both statements are correct and second statement is a correct explanation of statement - 1.