Permutation And Combination - Study Mode

[#71] A box contains 2 white, 3 black and 4 red balls. In how many ways can 3 balls be drawn from the box, if at least 1 black ball is to be included in the draw?
Correct Answer

(C) 64

Explanation

Solution: We may have (1 black and 2 non-black) or (2 black and 1 non-black) or (3 black). Required number of ways $$eqalign{
& left( {{}^3{C_1} imes {}^6{C_2}}
ight) + left( {{}^3{C_2} imes {}^6{C_1}}
ight) + left( {{}^3{C_3}}
ight) cr
& = left{ {3 imes frac{{6 imes 5}}{{2 imes 1}}}
ight} + left( {frac{{3 imes 2}}{{2 imes 1}} imes 6}
ight) + 1 cr
& = left( {45 + 18 + 1}
ight) cr
& = 64 cr} $$

[#72] Out of 5 men and 3 women, a committee of three members is to be formed so that it has 1 women, and 2 men. In how many different ways can it be done?
Correct Answer

(D) 30

Explanation

Solution: Required number of ways $$eqalign{
& = left( {{}^3{C_1} imes {}^5{C_2}}
ight) cr
& = 3 imes frac{{5 imes 4}}{{2 imes 1}} cr
& = 30 cr} $$

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

(A) 60

Explanation

Solution: The given words contains 6 letters of which A is taken 3 times, N is taken 2 times and the rest are all different. ∴ Required number of ways $$eqalign{
& = frac{{6!}}{{3! imes 2!}} cr
& = frac{{6 imes 5 imes 4 imes 3 imes 2 imes 1}}{{6 imes 2}} cr
& = 60 cr} $$

[#74] A committee of 5 members is to be formed out of 3 trainees, 4 professors and 6 research associates. In how many different ways can this be done, if the committee should have 4 professors and 1 research associate or all 3 trainees and 2 professors?
Correct Answer

22

Explanation

Solution: Required number of ways $$ = left( {{}^4{C_4} imes {}^6{C_1}}
ight) + $$ xa0 $$left( {{}^3{C_3} imes {}^4{C_{2}}}
ight)$$ $$eqalign{
& = left( {1 imes 6}
ight) + left( {1 imes frac{{4 imes 3}}{2}}
ight) cr
& = left( {6 + 6}
ight) cr
& = 12 cr} $$

[#75] How many arrangements of four 0's (zeroes), two 1's and two 2's are there in which the first 1 occur before the first 2?
Correct Answer

(D) 210

Explanation

Solution: Total number of arrangements = $$frac{{8!}}{{4! imes 2! imes 2!}}$$ xa0xa0= 420 Since, there are two 1's and two 0's, the number of arrangements in which the first 1 is before the first 2 is same as the number of arrangement in which the first 2 is before the first 1 and they are each equal to half the total number of arrangements = 210