Aptitude  Permutation and Combination
 Permutation and Combination  Formulas
 Permutation and Combination  General Questions
We may have (1 boy and 3 girls) or (2 boys and 2 girls) or (3 boys and 1 girl) or (4 boys).
Required number of ways 
= (^{6}C_{1} x ^{4}C_{3}) + (^{6}C_{2} x ^{4}C_{2}) + (^{6}C_{3} x ^{4}C_{1}) + (^{6}C_{4})  
= (^{6}C_{1} x ^{4}C_{1}) + (^{6}C_{2} x ^{4}C_{2}) + (^{6}C_{3} x ^{4}C_{1}) + (^{6}C_{2})  


= (24 + 90 + 80 + 15)  
= 209. 
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20.
Required number of ways  = (^{8}C_{5} x ^{10}C_{6})  
= (^{8}C_{3} x ^{10}C_{4})  


= 11760. 
We may have(1 black and 2 nonblack) or (2 black and 1 nonblack) or (3 black).
Required number of ways  = (^{3}C_{1} x ^{6}C_{2}) + (^{3}C_{2} x ^{6}C_{1}) + (^{3}C_{3})  


= (45 + 18 + 1)  
= 64. 
There are 6 letters in the given word, out of which there are 3 vowels and 3 consonants.
Let us mark these positions as under:
(1) (2) (3) (4) (5) (6)
Now, 3 vowels can be placed at any of the three places, marked 1, 3, 5.
Number of ways of arranging the vowels = ^{3}P_{3} = 3! = 6.
Also, the 3 consonants can be arranged at the remaining 3 positions.
Number of ways of these arrangements = ^{3}P_{3} = 3! = 6.
Total number of ways = (6 x 6) = 36.