The number of one-one functions f : a, b, c, d → 0, 1, 2, ......, 10 such
that 2f(a) - f(b) + 3f(c) + f(d) = 0 is ___________.

Solution

Maths sir · 3 steps

Rewrite it as f(b) = 2f(a) + 3f(c) + f(d); since f(b) is at most 10, f(c) can only be 0, 1, 2 or 3.

1. Isolate f(b). Let p = f(a), q = f(c), r = f(d). Then f(b) = 2p + 3q + r, and this must be at most 10. So 3q ≤ 10, which leaves q ∈ {0, 1, 2, 3}.
2. f(b) is automatically new. Since p, q, r are distinct, at most one of them is 0. So f(b) - p = p + 3q + r, f(b) - r = 2p + 3q and f(b) - q = 2p + 2q + r are all positive. That means f(b) never repeats a value. You only need p, q, r distinct and 2p + 3q + r ≤ 10.

   f(c) = q	Condition on (p, r)	Count, by value of p	Total
   0	2p + r ≤ 10, p ≠ r, neither is 0	p = 1: 7, p = 2: 5, p = 3: 3, p = 4: 2	17
   1	2p + r ≤ 7, p ≠ r, neither is 1	p = 0: 6, p = 2: 2, p = 3: 1	9
   2	2p + r ≤ 4, p ≠ r, neither is 2	p = 0: 3, p = 1: 1	4
   3	2p + r ≤ 1, p ≠ r, neither is 3	only p = 0, r = 1, giving f(b) = 10	1

3. So. The number of one-one functions is 17 + 9 + 4 + 1 = 31.

The idea to keep. When a one-one condition comes with an equation, put the variable with the largest coefficient in cases first. The range bound cuts it down to a few values, and each case becomes a short count.

Answer from the NTA final answer key. Solution by PrepWiser.