Permutation And Combination - Study Mode

[#141] There are 10 seats around a circular table. If 8 men and 2 women have to seated around a circular table, such that no two women have to be separated by at least one man. If P and Q denote the respective number of ways of seating these people around a table when seats are numbered and unnumbered, then P : Q equals
Correct Answer

(C) 10 : 1

Explanation

Solution: 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

[#142] How many factors of 2 5 × 3 6 × 5 2 are perfect squares?
Correct Answer

(B) 24

Explanation

Solution: 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

[#143] From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
Correct Answer

(D) 756

Explanation

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

[#144] In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
Correct Answer

(C) 720

Explanation

Solution: 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

[#145] In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?
Correct Answer

(D) 50400

Explanation

Solution: 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