Permutation And Combination - Study Mode

[#156] How many ways can 4 prizes be given away to 3 boys, if each boy is eligible for all the prizes?
Correct Answer

(B) 24

Explanation

Solution: Let the 3 boys be B 1 , B 2 , B 3 and 4 prizes be P 1 , P 2 , P 3 and P 4 Now B 1 is eligible to receive any of the 4 available prizes (so 4 ways)
B 2 will receive prize from rest 3 available prizes(so 3 ways)
B 3 will receive his prize from the rest 2 prizes available(so 2 ways)
So total ways would be: 4 × 3 × 2 × 1 = 24 Ways Hence, the 4 prizes can be distributed in 24 ways

[#157] Find the number of ways in which 8064 can be resolved as the product of two factors?
Correct Answer

(B) 24

Explanation

Solution: Total number of ways in which 8064 can be resolved as the product of two factors is 24 as below: (1,8064), (2,4032), (3,2688), (4,2016), (6,1344), (7,1152), (8,1008), (9,896), (12,672), (14,576), (16,504), (18,448), (21,884), (24,336), (28,288), (32,252), (36,224), (42,192), (48,168), (56,144), (63,128), (68,126), (72,112), (84,96)

[#158] Six boxes are numbered 1, 2, 3, 4, 5 and 6. Each box must contain either a white ball or a black ball. At least one box must contain a black ball and boxes containing black balls must be consecutively numbered. find the total number of ways of placing the balls.
Correct Answer

(C) 21

Explanation

Solution: If there is 1 black ball, it can be placed in 6 ways. If there are 2 black balls, they can be placed in 5 ways (in 1,2
2,3
3,4
4,5 and 5,6) and so on.
If there are 6 black balls, they can be placed in 1 way.
The total number of ways of placing the balls is = 1 + 2 + 3 + 4 + 5 + 6 = 21

[#159] In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys?
Correct Answer

(C) 25

Explanation

Solution: The toys are different
The boxes are identical. If none of the boxes is to remain empty, then we can pack the toys in one of the following ways: Case i. 2, 2, 1 Case ii. 3, 1, 1 Case i: Number of ways of achieving the first option 2, 2, 1 Two toys out of the 5 can be selected in 5 C 2 ways. Another 2 out of the remaining 3 can be selected in 3 C 2 ways and the last toy can be selected in 1 C 1 way. However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2. Therefore, total number of ways of achieving the 2, 2, 1 option is: $$frac{{^5{C_2}{ imes ^3}{C_2}}}{2} = frac{{10 imes 3}}{2} = 15,{ ext{ways}}$$ Case ii: Number of ways of achieving the second option 3, 1, 1 Three toys out of the 5 can be selected in $$^5{C_3}$$ ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way. Therefore, total number of ways of getting the 3, 1, 1 option is $$^5{C_3}$$ = 10 ways. Total ways in which the 5 toys can be packed in 3 identical boxes = number of ways of achieving Case i + number of ways of achieving Case ii = 15 + 10 = 25 ways.

[#160] There are 6 equally spaced points A, B, C, D, E and F marked on a circle with radius R. How many convex pentagons of distinctly different areas can be drawn using these points as vertices?
Correct Answer

(D) None of these
(H) None of these

Explanation

Solution: Since, all the points are equally spaced
hence the area of all the convex pentagons will be same.