Permutation And Combination - Aptitude MCQ Questions and Solutions with Explanations

There are six letters in the given word MOMENT and letter 'M' has come twice.
∴ Required number of ways
$$\eqalign{ & = \frac{{6!}}{{2!}} \cr & = \frac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1}} \cr & = 360 \cr} $$

There are 2 primary teachers.
They can stand in a row in
P (2, 2) = 2! = 2 × 1 ways = 2 ways
∴ Two middle teachers.
They can stand in a row in
P (2, 2) = 2! = 2 × 1 ways = 2 ways
There are two secondary teachers.
They can stand in a row in
P (2, 2) = 2!= 2 × 1 ways = 2 ways
These three sets can be arranged themselves in
3! ways = 3 × 2 × 1 = 6 ways
Hence,, the required number of ways
= 2 × 2 × 2 × 6
= 48 ways

The given word contains 8 different letters.
We keep the vowels (OAE) together and treat them as 1 letter.
Thus, we have to arrange the 6 letters SFTWR(OAE)
These can be arranged in 6! = 720 ways
The vowels (OAE) can be arranged among themselves in 3! = 6 ways.
∴ Required number of ways = (720 × 6) = 4320

The committee should have
(1 man, 4 women) or (2 men, 3 women) or (3 men, 2 women) or ( 4 men, 1 woman)
Required number of ways
$$ = \left( {{}^4{C_1} \times {}^5{C_4}} \right) + \left( {{}^4{C_2} \times {}^5{C_3}} \right)$$      $$ + \left( {{}^4{C_3} \times {}^5{C_2}} \right)$$   $$ + \left( {{}^4{C_6} \times {}^5{C_1}} \right)$$
$$ = \left( {{}^4{C_1} \times {}^5{C_1}} \right) + \left( {{}^4{C_2} \times {}^5{C_2}} \right)$$      $$ + \left( {{}^4{C_1} \times {}^5{C_2}} \right)$$   $$ + \left( {{}^4{C_4} \times {}^5{C_1}} \right)$$
$$ = \left( {4 \times 5} \right) + \left( {\frac{{4 \times 3}}{{2 \times 1}} \times \frac{{5 \times 4}}{{2 \times 1}}} \right)$$      $$ + \left( {4 \times \frac{{5 \times 4}}{{2 \times 1}}} \right)$$   $$ + \left( {1 \times 5} \right)$$
$$ = \left( {20 + 60 + 40 + 5} \right)$$
$$ = 125$$

Keeping the vowels (AEIA) together, we have MTHMTCS (AEAI).
Now, we have to arrange 8 letters, out of which we have 2M, 2T and the rest are all different.
Number of ways of arranging these letters
$$\eqalign{ & = \frac{{8!}}{{2!.2!}} \cr & = \frac{{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 2 \times 1}} \cr & = 10080 \cr} $$
Now, (AEAI) has 4 letters, out of which we have 2A, 1E and 1I.
Number of ways of arranging these letters
$$\eqalign{ & = \frac{{4!}}{{2!}} \cr & = \frac{{4 \times 3 \times 2 \times 1}}{2} \cr & = 12 \cr} $$
∴ Required number of ways = (10080 × 12) = 120960

The given word contains 5 letters of which T is taken 2 times.
∴ Required number of ways
$$\eqalign{ & = \frac{{5!}}{{2!}} \cr & = \frac{{5 \times 4 \times 3 \times 2!}}{{2!}} \cr & = 60 \cr} $$

The given word contains 5 letters, all different.
∴ Required number of ways
$$\eqalign{ & = {}^5{P_5} \cr & = 5! \cr & = \left( {5 \times 4 \times 3 \times 2 \times 1} \right) \cr & = 120 \cr} $$

$$\eqalign{ & = \left( {{}^{75}{P_2} - {}^{75}{C_2}} \right) \cr & = \left\{ {\frac{{75!}}{{75! - 2!}} - \frac{{75 \times 74}}{2}} \right\} \cr & = \frac{{75!}}{{73!}} - \left( {75 \times 37} \right) \cr & = \frac{{75 \times 74 \times 73!}}{{73!}} - \left( {75 \times 37} \right) \cr & = \left( {75 \times 74 - 75 \times 37} \right) \cr & = 75 \times 37 \times \left( {2 - 1} \right) \cr & = \left( {75 \times 37} \right) \cr & = 2775 \cr} $$

The given word contains 8 letters of which E is taken 3 times.
∴ Required number of ways
$$\eqalign{ & = \frac{{8!}}{{3!}} \cr & = \frac{{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{6} \cr & = 6720 \cr} $$

The given words contains 8 letters out of which U is taken 2 times and all other letters are different.
∴ Required number of ways
$$\eqalign{ & = \frac{{8!}}{{2!}} \cr & = \frac{{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{2} \cr & = 20160 \cr} $$