Permutation & Combination flash cards

1. 1.State the formula for the number of permutations of $n$ distinct objects taken $r$ at a time.

   $^nP_r=\dfrac{n!}{(n-r)!}$, where $0\le r\le n$.

   Hint: Arrangement — order matters.

2. 2.State the formula for $^nC_r$ and the key relation between $^nC_r$ and $^nP_r$.

   $^nC_r=\dfrac{n!}{r!(n-r)!}$ and $^nP_r={}^nC_r\times r!$.

   Hint: Selection — order doesn't matter.

3. 3.What is the number of ways to arrange $n$ objects in a circle, and how does it change if clockwise and anticlockwise arrangements are considered identical?

   Circular permutations $=(n-1)!$; if clockwise and anticlockwise arrangements are not distinct (as in arranging beads/flowers in a circle), it becomes $\dfrac{(n-1)!}{2}$.

   Hint: Fix one object to remove rotational symmetry.

4. 4.What is the number of distinguishable arrangements of $n$ objects when $p, q, r,\ldots$ of them are alike (of different kinds), and how many ways can $n$ distinct objects be divided into three groups of sizes $p, q, r$ (with $p+q+r=n$)?

   Arrangements with repetition: $\dfrac{n!}{p!\,q!\,r!\cdots}$; number of ways to divide $n$ distinct objects into groups of specified sizes $p,q,r$ is $\dfrac{n!}{p!\,q!\,r!}$ (this becomes $\dfrac{n!}{p!\,q!\,r!\,\times k!}$ if $k$ of the groups have the same size and are otherwise unmarked/identical).

   Hint: Permutations with repetition; watch for identical group sizes.

5. 5.State the useful identities $^nC_r+{}^nC_{r-1}$ and the sum $\sum_{r=0}^{n}{}^nC_r$, along with the number of non-negative integral solutions of $x_1+x_2+\cdots+x_r=n$.

   $^nC_r+{}^nC_{r-1}={}^{n+1}C_r$ (Pascal's rule); $\sum_{r=0}^{n}{}^nC_r=2^n$; number of non-negative integral solutions $={}^{n+r-1}C_{r-1}$.

   Hint: Pascal's rule and the stars-and-bars method.

6. 6.In how many ways can 8 distinct beads be arranged on a bracelet, where flipping the bracelet over gives the same arrangement?

   For a bracelet, both rotations and reflections give identical arrangements, so the count is $\dfrac{(n-1)!}{2}$. For $n=8$: $\dfrac{7!}{2}=\dfrac{5040}{2}=2520$.

   Hint: Bracelets have both rotational and reflective symmetry, unlike a plain circular table.

7. 7.5 couples (10 people) are seated around a circular table such that men and women alternate. Find the number of arrangements.

   Fix the men's circular arrangement first: $(5-1)!=24$ ways. The 5 alternate seats are then filled by the women in $5!=120$ ways. Total $=24\times120=2880$.

   Hint: Seat one gender in a circle first, then slot the other gender into the gaps.

8. 8.For $n$ distinct objects seated in a circle, find the number of arrangements in which two specified objects are never adjacent.

   Total circular arrangements $=(n-1)!$. Arrangements with the two objects together: glue them into one unit, giving $(n-2)!$ circular arrangements times $2!$ for their internal order, i.e. $2(n-2)!$. So the required count is $(n-1)!-2(n-2)!$.

   Hint: Subtract the 'glued together' circular arrangements from the total.

9. 9.6 men and 3 women are seated around a circular table so that no two women are adjacent. Find the number of arrangements.

   Seat the 6 men in a circle: $(6-1)!=120$ ways, creating 6 gaps between consecutive men. Choose 3 of these 6 gaps for the women, in order (since women are distinct): $P(6,3)=120$. Total $=120\times120=14400$.

   Hint: Fix the men's circle first, then place women in the gaps using a permutation, not a combination.

10. 10.12 people are split into two circular tables of 6 each, where the two tables are otherwise identical (unlabeled). Find the number of distinct seatings.

    Choose which 6 of the 12 sit at one table: $\binom{12}{6}=924$; since the tables are identical, divide by $2$. Each group of 6 is then seated circularly in $(6-1)!=120$ ways. Total $=\dfrac12\times924\times120\times120=6652800$.

    Hint: Divide by $2!$ to correct for the two tables being indistinguishable.

11. 11.4 Indian, 3 American, and 2 British delegates are seated around a round table so that delegates of the same nationality sit together. Find the number of arrangements.

    Treat each nationality as one block: 3 blocks arranged circularly in $(3-1)!=2$ ways. Within blocks: $4!\times3!\times2!=24\times6\times2=288$. Total $=2\times288=576$.

    Hint: Bundle each nationality into a single block before arranging circularly.

12. 12.8 people sit around a circular table with an even number of identical seats, so that A and B must sit diametrically opposite each other. Find the number of arrangements.

    Fix A's seat to remove rotational symmetry. B's seat is then forced (the unique seat opposite A), giving 1 way. The remaining 6 people fill the remaining 6 seats in $6!=720$ ways. Total $=720$.

    Hint: Fixing A's seat automatically fixes B's seat since it must be directly opposite.

13. 13.7 distinct flags are arranged in a circle so that one particular flag F must always sit exactly between two other particular flags G and H (in either order). Find the number of arrangements.

    Bundle F, G, H into a single block with F in the middle, giving 2 internal arrangements (G,H can swap). This block plus the remaining 4 flags gives 5 units arranged circularly: $(5-1)!=24$. Total $=24\times2=48$.

    Hint: Treat the fixed middle flag together with its two neighbours as one block.

14. 14.A key-ring holds $n$ distinct keys; arrangements are identical if related by rotation or by flipping the ring over. Compute the number of distinct arrangements for $n=6$.

    Since the ring can be flipped as well as rotated, distinct arrangements $=\dfrac{(n-1)!}{2}$. For $n=6$: $\dfrac{5!}{2}=\dfrac{120}{2}=60$.

    Hint: Flipping the ring is an extra symmetry beyond simple rotation.

15. 15.5 boys and 5 girls sit around a circular table so that boys and girls alternate. How many of these arrangements also have two particular girls not adjacent to each other?

    Alternating seating: fix boys circularly in $(5-1)!=24$ ways, then place girls in the alternate seats in $5!=120$ ways, giving $2880$ total. Since every girl is flanked by two boys in an alternating arrangement, no two girls can ever be adjacent — so the extra condition is automatically satisfied and the answer is still $\mathbf{2880}$.

    Hint: Check whether the alternating condition already forces the extra restriction to hold.

16. 16.How many distinct arrangements of the letters of 'MATHEMATICS' have all 4 vowels (A,A,E,I) together?

    Letters: M2,A2,T2,H,E,I,C,S. Bundle the 4 vowels into one block; remaining 7 consonants (M2,T2,H,C,S) plus the block form 8 units: $\dfrac{8!}{2!2!}=10080$ ways. The block's internal arrangements: $\dfrac{4!}{2!}=12$ (A repeats). Total $=10080\times12=120960$.

    Hint: Bundle the vowels, then handle repeated letters both inside and outside the block.

17. 17.How many arrangements of the letters of 'ARRANGEMENT' have no two R's adjacent?

    Letters: A2,R2,N2,E2,G,M,T (11 letters). Total arrangements $=\dfrac{11!}{2!2!2!2!}=2494800$. Arrangements with RR together: glue the R's, giving 10 units (A2,N2,E2,G,M,T,[RR]): $\dfrac{10!}{2!2!2!}=453600$. Required count $=2494800-453600=2041200$.

    Hint: Subtract the 'R's glued together' count (treated as one block) from the total.

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