The number of ways by which 6 distinct balls can be put in 5 distinct boxes are

Correct — B, 15625. Both the balls and the boxes are distinct, and nothing forbids a box from holding several balls or none. Take the balls one at a time: the first can go into any of the 5 boxes, and so can the second, the third and each of the rest, with no interference between the choices. Six independent choices from 5 possibilities each give 5 × 5 × 5 × 5 × 5 × 5 = 5⁶ = 15,625. The direction of the exponent is what the item is really testing. It is balls choosing boxes, not boxes choosing balls, because every ball must go somewhere while a box may stay empty — so the count is (number of boxes) raised to the (number of balls), and 5⁶ rather than 6⁵.

• (a)7776 — This is 6⁵, the exponent taken the wrong way round. It would count the ways of assigning a ball to each box, which is not what the question asks and would leave some balls unplaced.
• (c)720 — This is 6!, the number of ways of arranging six distinct objects in a row. There is no arrangement in this problem; the boxes are containers, not positions in a sequence.
• (d)120 — This is 5!, the arrangements of five objects. It would be relevant if the six balls had to be distributed one per box, which is impossible with six balls and five boxes.

Distribution problems split by two questions: are the objects distinct, and are the containers distinct? With distinct balls and distinct boxes and no restriction, the answer is simply (boxes)^(balls), because each ball is an independent choice. Add the condition that no box may be empty and the count becomes the number of onto functions, found by inclusion and exclusion. Make the balls identical instead and the count becomes a stars-and-bars combination.

The mistake this item punishes is inverting the exponent, and the guard against it is to ask which set has to be exhausted. Every ball must land in a box, so every ball makes a choice, and the number of choices each one has is the number of boxes. That reasoning gives 5⁶ every time and never 6⁵. The factorial options are there for candidates who see six objects and five containers and reach for an arrangement formula out of habit.

• With distinct balls and distinct boxes and no restriction, the count is (boxes) raised to the power (balls).
• Here that is 5⁶ = 15,625, not 6⁵ = 7,776.
• Each ball is placed independently, so the choices multiply.
• If no box were allowed to be empty the count would be the number of onto functions, found by inclusion and exclusion.
• If the balls were identical instead, the count would be a stars-and-bars combination rather than a power.

• Inverting the exponent and answering 6⁵.
• Reaching for a factorial because two counts of objects appear in the stem.
• Assuming implicitly that no box may be left empty when the question imposes no such condition.

Asked as a bare counting item whose four options are 6⁵, 5⁶, 6! and 5! — a design that gives no credit for a half-remembered formula.

• practice — not a real PYQ

  In how many ways can 4 distinct letters be put into 3 distinct post boxes?

  • (a)12
  • (b)24
  • (c)64
  • (d)81

  Answer(d) 81 — each of the 4 letters independently chooses one of the 3 boxes, giving 3⁴ = 81.

• practice — not a real PYQ

  In how many ways can 5 distinct balls be placed in 3 distinct boxes so that no box remains empty?

  • (a)60
  • (b)120
  • (c)150
  • (d)243

  Answer(c) 150 — 3⁵ − 3 × 2⁵ + 3 × 1⁵ = 243 − 96 + 3 = 150, by inclusion and exclusion.