Permutation And Combination - Study Mode

[#96] There are six teachers. Out of them two are primary teachers and two are secondary teachers. They are to stand in a row, so as the primary teachers, middle teachers and secondary teachers are always in a set . The number of ways in which they can do so, is-
Correct Answer

(B) 48

Explanation

Solution: 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

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

120

Explanation

Solution: 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

[#98] A committee of 5 members is to be formed by selecting out of 4 men and 5 women. In how many different ways the committee can be formed if it should have at least 1 man?
Correct Answer

(C) 125

Explanation

Solution: 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 $$ = left( {{}^4{C_1} imes {}^5{C_4}}
ight) + left( {{}^4{C_2} imes {}^5{C_3}}
ight)$$ xa0 xa0xa0 $$ + left( {{}^4{C_3} imes {}^5{C_2}}
ight)$$ xa0 $$ + left( {{}^4{C_6} imes {}^5{C_1}}
ight)$$ $$ = left( {{}^4{C_1} imes {}^5{C_1}}
ight) + left( {{}^4{C_2} imes {}^5{C_2}}
ight)$$ xa0 xa0xa0 $$ + left( {{}^4{C_1} imes {}^5{C_2}}
ight)$$ xa0 $$ + left( {{}^4{C_4} imes {}^5{C_1}}
ight)$$ $$ = left( {4 imes 5}
ight) + left( {frac{{4 imes 3}}{{2 imes 1}} imes frac{{5 imes 4}}{{2 imes 1}}}
ight)$$ xa0 xa0xa0 $$ + left( {4 imes frac{{5 imes 4}}{{2 imes 1}}}
ight)$$ xa0 $$ + left( {1 imes 5}
ight)$$ $$ = left( {20 + 60 + 40 + 5}
ight)$$ $$ = 125$$

[#99] In how many ways can the letters of the word MATHEMATICS be arranged so that all the vowels always come together?
Correct Answer

(B) 120960

Explanation

Solution: 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 $$eqalign{
& = frac{{8!}}{{2!.2!}} cr
& = frac{{8 imes 7 imes 6 imes 5 imes 4 imes 3 imes 2 imes 1}}{{2 imes 1 imes 2 imes 1}} cr
& = 10080 cr} $$ Now, (AEAI) has 4 letters, out of which we have 2A, 1E and 1I. Number of ways of arranging these letters $$eqalign{
& = frac{{4!}}{{2!}} cr
& = frac{{4 imes 3 imes 2 imes 1}}{2} cr
& = 12 cr} $$ ∴ Required number of ways = (10080 × 12) = 120960

[#100] In how many different ways can the letters of the word TOTAL be arranged?
Correct Answer

(B) 60

Explanation

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