Permutation And Combination - Aptitude MCQ Questions and Solutions with Explanations

If n is even, then the number of boys should be equal to number of girls, let each be a.
⇒ n = 2a
Then the number of arrangements = 2 × a! × a!
If one more students is added, then number of arrangements,
= a! × (a + 1)!
But this is 200% more than the earlier
⇒ 3 × (2 × a! × a!) = a! × (a + 1)!
⇒ a + 1 = 6 and a = 5
⇒ n = 10
But if n is odd, then number of arrangements, = a!(a + 1)!
Where, n = 2a + 1
When one student is included, number of arrangements,
= 2(a + 1)! (a + 1)!
By the given condition, 2(a + 1) = 3, which is not possible.

The smallest number in the series is 1000, a 4-digit number.
The largest number in the series is 4000, the only 4-digit number to start with 4.
The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3.
The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4.
Hence, there are 3 × 5 × 5 × 5 or 375 numbers from 1000 to 3999
Including 4000, there will be 376 such numbers.

Test of divisibility for 3:
The sum of the digits of any number that is divisible by 3 is divisible by 3
For instance, take the number 54372
Sum of its digits is 5 + 4 + 3 + 7 + 2 = 21
As 21 is divisible by 3, 54372 is also divisible by 3
There are six digits viz., 0, 1, 2, 3, 4 and 5. To form 5-digit numbers we need exactly 5 digits. So we should not be using one of the digits.

The sum of all the six digits 0, 1, 2, 3, 4 and 5 is 15. We know that any number is divisible by 3 if and only if the sum of its digits is divisible by 3

Combining the two criteria that we use only 5 of the 6 digits and pick them in such a way that the sum is divisible by 3, we should not use either 0 or 3 while forming the five digit numbers.

Case 1
If we do not use '0', then the remaining 5 digits can be arranged in:
5! ways = 120 numbers.

Case 2
If we do not use '3', then the arrangements should take into account that '0' cannot be the first digit as a 5-digit number will not start with '0'.

The first digit from the left can be any of the 4 digits 1, 2, 4 or 5
Then the remaining 4 digits including '0' can be arranged in the other 4 places in 4! ways.

So, there will be 4 × 4! numbers = 4 × 24 = 96 numbers.

Combining Case 1 and Case 2, there are a total of 120 + 96 = 216, 5 digit numbers divisible by '3' that can be formed using the digits 0 to 5.

Initially we look at the general case of the seats not numbered.
The total number of cases of arranging 8 men and 2 women, so that women are together,
⇒ 8! ×2!

The number of cases where in the women are not together,
⇒ 9! - (8! × 2!) = Q

Now, when the seats are numbered, it can be considered to a linear arrangement and the number of ways of arranging the group such that no two women are together is,
⇒ 10! - (9! × 2!)

But the arrangements where in the women occupy the first and the tenth chairs are not favorable as when the chairs which are assumed to be arranged in a row are arranged in a circle, the two women would be sitting next to each other.

The number of ways the women can occupy the first and the tenth position,
= 8! × 2!

The value of P = 10! - (9! × 2!) - (8! × 2!)
Thus P : Q = 10 : 1

Any factor of this number should be of the form 2^a × 3^b × 5^c
For the factor to be a perfect square a, b, c have to be even.
a can take values 0, 2, 4, b can take values 0, 2, 4, 6 and c can take values 0, 2
Total number of perfect squares =3 × 4 × 2 = 24

We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
∴ the Required number of ways
$$\eqalign{ & = \left( {{}^7{C_3} \times {}^6{C_2}} \right) + \left( {{}^7{C_4} \times {}^6{C_1}} \right) + \left( {{}^7{C_5}} \right) \cr & = \left( {\frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \times \frac{{6 \times 5}}{{2 \times 1}}} \right) + \left( {{}^7{C_3} \times {}^6{C_1}} \right) + \left( {{}^7{C_2}} \right) \cr & = 525 + \left( {\frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \times 6} \right) + \left( {\frac{{7 \times 6}}{{2 \times 1}}} \right) \cr & = \left( {525 + 210 + 21} \right) \cr & = 756 \cr} $$

The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Therefore Required number of ways = (120 x 6) = 720

In the word 'CORPORATION', we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters = $$\frac{{7!}}{{2!}}$$ = 2520
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in $$\frac{{5!}}{{3!}}$$ = 20 ways
∴ Required number of ways = (2520 x 20) = 50400

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
$$\eqalign{ & = \left( {{}^7{C_3} \times {}^4{C_2}} \right) \cr & = \left( {\frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \times \frac{{4 \times 3}}{{2 \times 1}}} \right) \cr & = 210 \cr} $$
Number of groups, each having 3 consonants and 2 vowels = 210
Each group contains 5 letters.
Number of ways of arranging 5 letters among themselves
= 5!
= 5 x 4 x 3 x 2 x 1
= 120
∴ Required number of ways = (210 x 120) = 25200

The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
∴ Required number of ways $$ = \frac{{6!}}{{\left( {1!} \right)\left( {2!} \right)\left( {1!} \right)\left( {1!} \right)\left( {1!} \right)}} = 360$$