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Permutations and Combinations questions for IPMAT, with solutions

10 free Permutations and Combinations practice questions for IPMAT Indore, Rohtak and JIPMAT, 5 from past papers, each with the answer and a full worked solution. No sign-in. Try each one first, then open the solution. The lessons for this chapter are in Permutations and Combinations on Learn.

1. Question 1Multiple choiceAsked in IPMAT Rohtak 2020Arrangements

   Can you calculate in how many ways can $7$ tennis players can be seated in a circular order?

   1. A.$330$
   2. B.$730$
   3. C.$720$
   4. D.$820$
   Show the answer and solution

   Answer: C. $720$

   Fix one player to remove rotations, then arrange the other $6$ in a line. $(7-1)!=6!=720$. Option C.

2. Question 2Multiple choiceAsked in IPMAT Indore 2021Counting in Geometry and Grids

   There are $10$ points in the plane, of which $5$ points are collinear and no three among the remaining are collinear. Then the number of distinct straight lines that can be formed out of these $10$ points is

   1. A.$10$
   2. B.$25$
   3. C.$35$
   4. D.$36$
   Show the answer and solution

   Answer: D. $36$

   Cases: the collinear points give $1$ line; the other $5$ give $\binom{5}{2}=10$; mixed pairs give $5\times5=25$. Total $36$.

3. Question 3Multiple choiceAsked in IPMAT Rohtak 2020Selections

   Please calculate in how many ways can a set of five players be formed out of a total of ten players such that two particular players should be involved in each set?

   1. A.$60$
   2. B.$72$
   3. C.$56$
   4. D.$75$
   Show the answer and solution

   Answer: C. $56$

   Two particular players are already in, so choose the other $3$ from the remaining $8$. $\binom{8}{3}=\frac{8\cdot7\cdot6}{6}=56$. Option C.

4. Question 4Type the answerWritten by IPMATwayArrangements

   A host and a hostess invite $6$ guests to dinner at a round table with $8$ equally spaced seats. If the host and hostess must sit directly opposite each other, in how many ways can the $8$ people be seated?

   Show the answer and solution

   Answer: 720

   Seating around a table counts arrangements up to rotation, so fix the host's seat. The hostess must take the opposite seat (one way), and the guests can be arranged in $6!=720$ ways.

5. Question 5Type the answerWritten by IPMATwayDistribution and Partitions

   How many non-negative integer solutions does $x+y+z=10$ have with $x\le 3$?

   Show the answer and solution

   Answer: 38

   Or: all $\binom{12}{2}=66$ minus those with $x\ge4$: $\binom{8}{2}=28$, giving $38$.

6. Question 6Multiple choiceAsked in IPMAT Indore 2025Counting Numbers and Words

   The number of integers greater than $5000$ and divisible by $5$ that can be formed with the digits $1, 3, 5, 7, 8, 9$ where no digit is repeated is

   1. A.$180$
   2. B.$120$
   3. C.$276$
   4. D.$240$
   Show the answer and solution

   Answer: C. $276$

   Every $5$- and $6$-digit number exceeds $5000$. A $4$-digit number ending in $5$ exceeds $5000$ only if it starts with $7$, $8$ or $9$ ($5$ is already used at the end): $3\times4\times3=36$. Adding $120+120$ gives $276$.

7. Question 7Multiple choiceWritten by IPMATwayCounting Numbers and Words

   How many integers from $1$ to $1000$ (both included) do not contain the digit $7$?

   1. A.$271$
   2. B.$728$
   3. C.$648$
   4. D.$729$
   Show the answer and solution

   Answer: D. $729$

   Strings $000$ to $999$ with no $7$: $9^3=729$; remove $000$ (not in range) and add $1000$ (no $7$): $729$.

8. Question 8Multiple choiceAsked in IPMAT Rohtak 2020Distribution and Partitions

   Five racquets needs to be placed in three boxes. Each box can hold all the five racquets. In how many ways can the racquets be placed in the boxes so that no box can be empty if all racquets are different but all boxes are identical?

   1. A.$24$
   2. B.$25$
   3. C.$27$
   4. D.$26$
   Show the answer and solution

   Answer: B. $25$

   With identical boxes only the grouping matters. $(3,1,1)$: choose the three together, $10$ ways. $(2,2,1)$: choose the single racquet ($5$ ways), split the other four into two pairs ($3$ ways): $15$. Total $25$ (the Stirling number $S(5,3)$).

9. Question 9Multiple choiceWritten by IPMATwayCounting in Geometry and Grids

   A rectangle is divided into a $5\times4$ grid of unit squares (5 columns, 4 rows). How many rectangles that are not squares can be found along the grid lines?

   1. A.$110$
   2. B.$150$
   3. C.$40$
   4. D.$60$
   Show the answer and solution

   Answer: A. $110$

   The $5\times4$ grid has $6$ vertical and $5$ horizontal grid lines. Rectangles: choose two of each, $\binom62\binom52=15\times10=150$. Squares of side $k$: $(6-k)(5-k)$ for $k=1..4$: $20+12+6+2=40$. Non-square rectangles $=150-40=110$. The trap is counting cells ($5$ and $4$) instead of lines ($6$ and $5$).

10. Question 10Multiple choiceWritten by IPMATwayTournaments and Handshakes

    In a round-robin tournament every pair of players was to play once. Two players withdrew after each had played exactly $3$ matches (and they had not played each other). The matches involving them still count. If a total of $84$ matches were played, how many players were there originally?

    1. A.$13$
    2. B.$14$
    3. C.$17$
    4. D.$15$
    Show the answer and solution

    Answer: D. $15$

    The $n-2$ players who stayed played all matches among themselves: $\binom{n-2}{2}$. The two who withdrew played $3+3=6$ matches, none against each other and none counted twice. Total $\binom{n-2}2+6=84$, so $\binom{n-2}{2}=78=\frac{13\cdot12}2$, giving $n-2=13$ and $n=15$.