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
a. 2,2,1
b. 3,1,1

Case a. Number of ways of achieving the first option 2−2–1

Two toys out of the 5 can be selected in5C2 5 ways. Another 2 out of the remaining 3 can be selected in3C2 ways and the last toy can be selected in 1C1  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:

5C2*3C22=10*32=15 ways.

Case b. Number of ways of achieving the second option 3−1–1

Three toys out of the 5 can be selected in 5C3  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 is5C3=10 ways.

Total ways in which the 5 toys can be packed in 3 identical boxes
=number of ways of achieving Case a + number of ways of achieving Case b
=15 + 10 = 25 ways

Let the two parallel sides of the trapezium be a cm and b cm. 

Then, a - b = 4 

And, 12×a+b×19=475=>a+b=50 

Solving (i) and (ii), we get: a = 27, b = 23. 

So, the two parallel sides are 27 cm and 23 cm.

Let breadth = x metres, length = 3x metres, height = H metres. 

Area of the floor=(Total cost of carpeting)/(Rate) = (270/5) sq.m = 54 sq.m 

x×3x2=54⇒x2=54×2⇒x=6  

So, breadth = 6 m and length =362 = 9 m.

Now, papered area = (1720/10) =  172 sq.m 

Area of 1 door and 2 windows = 8 sq.m 

Total area of 4 walls = (172 + 8) sq.m = 180 sq.m 

2×9+6H=180⇒H=6

This is the number of permutations of five things taken all at a time.

Therefore, answer = 5P5 = 120 ways

Let the side of the square(ABCD) be x meters.

Then, AB + BC = 2x metres.

AC = 2x = (1.41x) m.

Saving on 2x metres = (0.59x) m.

Saving % =0.59x2x*100 = 30% (approx)

Maximim number of triggers that can be applied to a single table is 12

No.Minimal spanning tree assures that the total weight of the tree is kept at its minimum. But it doesn’t mean that the distance between any two nodes involved in the minimum-spanning tree is minimum.

LDI is the capacity to change the conceptual schema without having to change external schemas or application programs.