Permutation And Combination - Aptitude MCQ Questions and Solutions with Explanations

We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
∴ Required number of ways
$$ = \left( {^6{C_1}{ \times ^4}{C_3}} \right) + \left( {^6{C_2}{ \times ^4}{C_2}} \right) + $$      $$\left( {^6{C_3}{ \times ^4}{C_1}} \right) + $$   $$\left( {^6{C_4}} \right)$$
$$ = \left( {^6{C_1}{ \times ^4}{C_1}} \right) + \left( {^6{C_2}{ \times ^4}{C_2}} \right) + $$      $$\left( {^6{C_3}{ \times ^4}{C_1}} \right) + $$   $$\left( {^6{C_2}} \right)$$
$$ = \left( {6 \times 4} \right) + \left( {\frac{{6 \times 5}}{{2 \times 1}} \times \frac{{4 \times 3}}{{2 \times 1}}} \right) + $$       $$\left( {\frac{{6 \times 5 \times 4}}{{3 \times 2 \times 1}} \times 4} \right) + $$    $$\left( {\frac{{6 \times 5}}{{2 \times 1}}} \right)$$
$$\eqalign{ & = \left( {24 + 90 + 80 + 15} \right) \cr & = 209 \cr} $$

Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
∴ Required number of numbers = (1 x 5 x 4) = 20

$$\eqalign{ & {\text{Required}}\,{\text{number}}\,{\text{of}}\,{\text{ways}} \cr & = {{}^8{C_5} \times {}^{10}{C_6}} \cr & = {{}^8{C_3} \times {}^{10}{C_4}} \cr & = {\frac{{8 \times 7 \times 6}}{{3 \times 2 \times 1}} \times \frac{{10 \times 9 \times 8 \times 7}}{{4 \times 3 \times 2 \times 1}}} \cr & = 11760 \cr} $$

We may have(1 black and 2 non-black) or (2 black and 1 non-black) or (3 black).
∴ Required number of ways
$$\eqalign{ & = \left( {{}^3{C_1} \times {}^6{C_2}} \right) + \left( {{}^3{C_2} \times {}^6{C_1}} \right) + \left( {{}^3{C_3}} \right) \cr & = \left( {3 \times \frac{{6 \times 5}}{{2 \times 1}}} \right) + \left( {\frac{{3 \times 2}}{{2 \times 1}} \times 6} \right) + 1 \cr & = \left( {45 + 18 + 1} \right) \cr & = 64 \cr} $$

There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.
Let us mark these positions as under:
(1) (2) (3) (4) (5) (6)
Now, 3 vowels can be placed at any of the three places out 4, marked 1, 3, 5
Number of ways of arranging the vowels = ³P₃ = 3! = 6
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = ³P₃ = 3! = 6
Total number of ways = (6 x 6) = 36

$$\eqalign{ & {\text{Required number of ways}} \cr & = {{}^7{C_5} \times {}^3{C_2}} \cr & = {{}^7{C_2} \times {}^3{C_1}} \cr & = {\frac{{7 \times 6}}{{2 \times 1}} \times 3} \cr & = 63 \cr} $$

'LOGARITHMS' contains 10 different letters.
Required number of words
= Number of arrangements of 10 letters, taking 4 at a time.
= ¹⁰P₄
= (10 x 9 x 8 x 7)
= 5040

In the word 'MATHEMATICS', we treat the vowels AEAI as one letter.
Thus, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which M occurs twice, T occurs twice and the rest are different.
∴ Number of ways of arranging these letters = $$\frac{{8!}}{{\left( {2!} \right)\left( {2!} \right)}}$$   = 10080
Now, AEAI has 4 letters in which A occurs 2 times and the rest are different.
Number of ways of arranging these letters = $$\frac{{4!}}{{2!}}$$ = 12
∴ Required number of words = (10080 x 12) = 120960

The word 'OPTICAL' contains 7 different letters.
When the vowels OIA are always together, they can be supposed to form one letter.
Then, we have to arrange the letters PTCL (OIA).
Now, 5 letters can be arranged in 5! = 120 ways.
The vowels (OIA) can be arranged among themselves in 3! = 6 ways.
Therefore Required number of ways = (120 x 6) = 720

Choosing 2 people out of 20 will result in a handshake and the same can be done in ²⁰C₂ ways
$${ \Rightarrow ^{20}}{C_2} = \frac{{20 \times 19}}{{2!}} = 190$$