Permutation and Combination Quiz – 1

Required number of ways =^{7}C_{5}\times^{3}C_{2}=^{7}C_{2}\times^{3}C_{1} = \frac{7\times6}{2\times1}\times 3 = 63

the word 'OPTICAL' contains 7 different letters when the vowels OIA are always together, they can be supposed to form one letter then, we have to arrange the letters PTCL(OIA) now, 5 letters can be arranged in 5! ways = 120 ways the vowels OIA can be arranged themselves in 3! ways = 6 ways required number of ways = 120 * 6 = 720 ways

Let n be the number of teams.
^{n}C_{2}=21
\frac{n(n-1)}{2}=21
or, n(n-1) = 42
or, n = 7

The candidate will fail if he fails either in 1 or 2 or 3 or 4 or 5 subjects, therefore, Required number of ways ^{5}C_{1}+^{5}C_{2}+^{5}C_{3}+^{5}C_{4}+^{5}C_{5}=31

Required number of ways = ^{15}C_{11}=^{15}C_{15-11}=^{15}C_{4}=1365

Required number of ways =^{7}C_{5}\times^{3}C_{2}=^{7}C_{2}\times^{3}C_{1} = \frac{7\times6}{2\times1}\times 3 = 63

Selecting 3 girls from 5 girls can be done in ^{5}C_{3} = 10 ways. Each of the boys may or may not be selected, therefore 16 ways. hence, 16*10 = 160 ways

Total possible ways = ^{24}C_{2} = 276

If students are A, B, C, D, E and F; we can have ^{6}C_{3} groups in all. However, if we have to count groups in which a particular student (say A) is always selected we would get ^{5}C_{2} = 10 ways of doing it.

For one pair of black shoes we require one left black and one right black.
Consider the worst case situation:
7LB + 5LW + 5RW + 1RB (or) 7RB + 5LW + 5RW + 1LB = 18 Shoes