Week 6 Graded Assignment
Statistics Graded Assignment
1. Five-Digit Numbers Divisible by 4
Question: How many 5-digit numbers can be formed from the numbers 0, 2, 4, 5, 7 and 9 (without repetition), such that it is divisible by 4?
Options: a. 120 b. 144 c. 132 d. 104
Answer: b. 144
Solution: A number is divisible by 4 if its last two digits form a number divisible by 4. Possible last two digits: 04, 20, 24, 40, 52, 72, 92.
- Case 1: Ends with 04, 20, or 40
- First three digits from remaining four digits: $^4P_3 = 24$ ways for each.
- Case 2: Ends with 24, 52, 72, or 92
- First digit cannot be 0.
- First digit: 3 choices (excluding 0 and the two used in the last two digits).
- Next two digits: from the remaining three (including 0), $^3P_2 = 6$ ways.
- Total for each: $3 \times 6 = 18$.
Total: $3 \times 24 + 4 \times 18 = 72 + 72 = 144$.
2. Train Tickets Between Stations
Question: There are $n$ train stops between Chennai and Assam. How many train tickets are to be printed, so that a person can travel between any of the two stations (irrespective of direction of travel)?
Answer: $(n+2)(n+1)$
Solution: Total stations = $n+2$ (including Chennai and Assam). Number of ways to select any two stations: $^{(n+2)}P_2 = (n+2)(n+1)$.
Example: If $n=7$, total tickets = $9 \times 8 = 72$.
3. Selecting Friends for a Party
Question: A man desires to throw a party for some of his friends. In how many ways can he select $m$ friends from a group of $n$ friends, if two of his friends (say ‘A’ and ‘B’) will not attend the party together?
Answer: $\binom{n}{m} - \binom{n-2}{m-2}$
Solution: Total ways without restriction: $\binom{n}{m}$. Ways where both A and B are present: $\binom{n-2}{m-2}$. Ways where A and B are not both present: $\binom{n}{m} - \binom{n-2}{m-2}$.
Example: If $n=10$, $m=7$, $\binom{10}{7} - \binom{8}{5} = 120 - 56 = 64$.
4. Wearing Clothes on Different Days
Question: Suman has $m$ clothes of different types, say, $C_1, C_2, …, C_m$ and she wants to wear all these clothes at different days, say, $D_1, D_2, …, D_m$. Due to some reason, $C_1$ must be used either at $D_{m-2}$ or at $D_{m-1}$ and $C_2$ can be used either at $D_{m-1}$ or at $D_{m-2}$ and $D_m$. Every cloth is to be used at only one day. In how many ways can clothes be used?
Answer: $2 \times 2 \times (m-2)!$
Solution:
- $C_1$: 2 ways ($D_{m-2}$ or $D_{m-1}$).
- $C_2$: 2 ways (if $C_1$ is used at $D_{m-1}$, $C_2$ can be used at $D_{m-2}$ or $D_m$; if $C_1$ is at $D_{m-2}$, $C_2$ at $D_{m-1}$ or $D_m$. However, the solution simplifies to 2 ways for $C_2$ as per the PDF.)
- Remaining clothes: $(m-2)$ clothes to be assigned to remaining days: $(m-2)!$ ways.
Total: $2 \times 2 \times (m-2)!$
Example: If $m=6$, total ways = $2 \times 2 \times 4! = 4 \times 24 = 96$.
5. Palindromic Numbers
Question: How many $n$-digit numbers can be formed such that they read the same way from either side (i.e., the number should be a palindrome)?
Options: a. $9 \times 10^{\lfloor \frac{n-1}{2} \rfloor}$ b. $9 \times 10^{\lceil \frac{n-1}{2} \rceil}$ c. $9 \times 10^{n-1}$ d. $10^n$
Answer: a. $9 \times 10^{\lfloor \frac{n-1}{2} \rfloor}$
Solution:
- First digit: 9 choices (cannot be 0).
- Next $\lfloor \frac{n-1}{2} \rfloor$ digits: 10 choices each.
- Rest: Determined by symmetry.
Total: $9 \times 10^{\lfloor \frac{n-1}{2} \rfloor}$.
6. Forming $m$-Digit Numbers Without Repetition
Question: Find the total number of ways to form an $m$-digit number (without repetition) from the digits $0, 1, 2, …, n$.
Options: a. $n \times {}^nP_{m-1}$ b. $^{n+1}P_m$ c. $(n-1) \times {}^{n-1}P_{m-1}$ d. $n + {}^nP_{m-1}$
Answer: a. $n \times {}^nP_{m-1}$
Solution:
- First digit: $n$ choices (cannot be 0).
- Remaining $m-1$ digits: $^nP_{m-1}$ (from the remaining $n$ digits).
Total: $n \times {}^nP_{m-1}$.
7. Round Table Seating
Question: In a restaurant, $x$ men and $y$ women are seated on $(x+y)$ chairs at a round table. Find the total number of possible ways such that $x$ men are always sitting next to each other.
Options: a. $x! \times y!$ b. $(x-1)! \times (y-1)!$ c. $x! \times (y+1)!$ d. $(x+y-1)!$
Answer: a. $x! \times y!$
Solution:
- Treat men as one group: $(y+1)$ entities, arrange in $y!$ ways (for round table, treating the group as fixed, but the PDF solution is $y! \times x!$).
- Arrange men within the group: $x!$ ways.
Total: $y! \times x!$ (as per PDF solution).
8. Group Formation with Exactly One District Level Player
Question: In how many ways can a group of $n-m$ players be formed from $n$ state level players and $m$ district level players such that the group contains exactly 1 district level player?
Answer: $\binom{n}{n-m-1} \times \binom{m}{1} = \frac{m \times n!}{(n-m-1)! \times (m+1)!}$
Solution:
- State level players: $\binom{n}{n-m-1}$
- District level player: $\binom{m}{1}$
- Total: $\binom{n}{n-m-1} \times \binom{m}{1}$
Example: If $n=10$, $m=6$, $\binom{10}{3} \times \binom{6}{1} = 120 \times 6 = 720$.
9. Permutation Ratio
Question: Find the value of $r$ such that the ratio of ${}^3P_r$ and ${}^4P_{r-1}$ will be $\frac{1}{2}$?
Answer: 3
Solution:
$$ \frac{3!/(3-r)!}{4!/(5-r)!} = \frac{1}{2} $$Solving: $(5-r)(4-r) = 2 \implies r = 3$ (since $r=6$ is invalid).
10. Incorrect Option for Combinations
Question: Choose the incorrect option/s for $n > 2$:
Options: a. $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$ b. $\binom{n}{r} = 1$ for $r = 0$ and $r = n$ c. $\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$ d. None of the above
Answer: d. None of the above
Solution: All options a, b, and c are correct for $n > 2$. Hence, the incorrect option is (d).
This covers all questions and solutions from the PDF1.
Week_6_Graded_Solution.pdf ↩︎