Permutation And Combination - Aptitude MCQ Questions and Solutions with Explanations

The given word contains 11 letters, namely 3E, 3N, 2G, 2I and 1R
So, Required number of ways :
$$\eqalign{ & = \frac{{11!}}{{3! \,3! \,2! \,2! \,1!}} \cr & = \frac{{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{6 \times 6 \times 2 \times 2 \times 1}} \cr & = \left( {11 \times 10 \times 9 \times 8 \times 7 \times 5} \right) \cr & = 277200 \cr} $$

Keeping the vowels (OOAIO) together as one letter we have CRPRTN (OOAIO).
This has 7 letters, out of which we have 2R, 1C, 1P, 1T and 1N.
Number of ways of arranging three letters
$$\eqalign{ & = \frac{{7!}}{{2!}} \cr & = \frac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} \cr & = 2520 \cr} $$
Now, (OOAIO) has 5 letters, out of which we have 3O, 1A and 1I.
Number of ways of arranging these letters
$$\eqalign{ & = \frac{{5!}}{{3!}} \cr & = \frac{{5 \times 4 \times 3 \times 2 \times 1}}{{3 \times 2 \times 1}} \cr & = 20 \cr} $$
∴ Required number of ways = (2520 × 20) = 50400

The given word contains 5 different letters.
Keeping the vowels UE together, we suppose them as 1 letter.
Then, we have to arrange the letters JDG (UE).
Now, we have to arrange in 4! = 24 ways.
The vowels (UE) can be arranged among themselves in 2 ways.
∴ Required number of ways = (24 × 2) = 48

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

The given word contains 7 letters, all different .
∴ Required number of ways
$$\eqalign{ & {}^7{P_7} = 7! \cr & = \left( {7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} \right) \cr & = 5040 \cr} $$

The given word contains 7 letters of which D is taken 3 times.
∴ Required number of ways
$$\eqalign{ & = \frac{{7!}}{{3!}} \cr & = \frac{{7 \times 6 \times 5 \times 4 \times 3!}}{{3!}} \cr & = \left( {7 \times 6 \times 5 \times 4} \right) \cr & = 840 \cr} $$

The given word contains 7 letters which D is taken 2 times.
∴ Required number of ways
$$\eqalign{ & = \frac{{7!}}{{2!}} \cr & = \frac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} \cr & = 2520 \cr} $$

Required number of ways
$$ = \left( {{}^6{C_1} \times {}^8{C_3}} \right) + \left( {{}^6{C_2} \times {}^8{C_2}} \right)$$     $$ + \left( {{}^6{C_3} \times {}^8{C_1}} \right)$$   $$ + \left( {{}^6{C_4} \times {}^8{C_0}} \right)$$
$$ = \left\{ {6 \times \frac{{8 \times 7 \times 6}}{{3 \times 2 \times 1}}} \right\} + $$    $$\left( {\frac{{6 \times 5}}{{2 \times 1}} \times \frac{{8 \times 7}}{{2 \times 1}}} \right)$$   $$ + \left( {\frac{{6 \times 5 \times 4}}{{3 \times 2 \times 1}} \times 8} \right)$$    $$ + \left( {{}^6{C_2} \times 1} \right)$$
$$ = \left\{ {6 \times \frac{{8 \times 7 \times 6}}{{3 \times 2 \times 1}}} \right\}$$    $$ +\, 420\, + $$  $$\left( {\frac{{6 \times 5 \times 4}}{6} \times 8} \right)$$   $$ + \left( {\frac{{6 \times 5}}{{2 \times 1}} \times 1} \right)$$
$$ = \left( {336 + 420 + 160 + 15} \right)$$
$$ = 931$$

The given word contains 7 different letters.
Keeping the vowels (AUIO) together, we take them as 1 letter.
Then, we have to arrange the letters CTN (AUIO).
Now, 4 letters can be arranged in 4! = 24 ways.
The vowels (AUIO) can be arranged among themselves in 4! = 24 ways.
∴ Required number of ways = (24 × 24) = 576

There are 7 letters in the given word, out of which there are 3 vowels and 4 consonants.
Let us mark the positions to be filled up as follows:
$$\left( {\mathop {}\limits^1 } \right)\left( {\mathop {}\limits^2 } \right)\left( {\mathop {}\limits^3 } \right)\left( {\mathop {}\limits^4 } \right)\left( {\mathop {}\limits^5 } \right)\left( {\mathop {}\limits^6 } \right)\left( {\mathop {}\limits^7 } \right)$$
Now, 3 vowels can placed at any of the three places out of four marked 1, 3, 5, 7
Number of ways of arranging the vowels
$$\eqalign{ & = {}^4{P_3} \cr & = \left( {4 \times 3 \times 2} \right) \cr & = 24 \cr} $$
4 consonants at the remaining 4 positions may be arranged in $${}^4{P_4} = 4! = $$   24 ways
Required number of ways = (24 × 24) = 576