IPMAT Indore 2024 · Quantitative Ability · Question 34
IPMAT Indore 2024 previous year question on Permutation and Combination, with the answer and a step-by-step solution.
- A5
- B6
- C2
- D4
Answer & solution
Answer: Option A
Step 1 — state when {}^{n}{C}_{r} is defined. We need
n\ge 0, r\ge 0, r\le n
with n and r non-negative integers. Here n=17-x and r=3x+1, so x must be an integer.
Step 2 — write the three inequalities.
17-x\ge 0 \Rightarrow x\le 17
3x+1\ge 0 \Rightarrow x\ge -1/3 \Rightarrow x\ge 0 \left( x\text{integer}\right)
3x+1\le 17-x \Rightarrow 4x\le 16 \Rightarrow x\le 4
Step 3 — intersect. The binding constraints are 0\le x\le 4:
x\in \{0,1,2,3,4\}
Step 4 — verify each.
x | n=17-x | r=3x+1 | {}^{n}{C}_{r} |
|---|---|---|---|
0 | 17 | 1 | {}^{17}{C}_{1} ✔ |
1 | 16 | 4 | {}^{16}{C}_{4} ✔ |
2 | 15 | 7 | {}^{15}{C}_{7} ✔ |
3 | 14 | 10 | {}^{14}{C}_{10} ✔ |
4 | 13 | 13 | {}^{13}{C}_{13} ✔ |
Answer: Option 1 — 5
Shortcut. Only one inequality does any work. r\ge 0 and n\ge 0 are nearly free; the binding condition is always
r\le n \Rightarrow 3x+1\le 17-x \Rightarrow x\le 4
Combined with x\ge 0 that is \{0,1,2,3,4\} — count 5.
Two habitual slips: forgetting that x=0 is allowed (giving 4, option d), and forgetting r\le n altogether (giving 18 values). Check the endpoints explicitly — here x=4 gives {}^{13}{C}_{13}=1, which is perfectly legitimate.
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