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- FSc Part 1 Mathematics Notes/Solutions @fsc
- SC-I), Punjab Textbook Board (PTB) Lahore.</lead> There are fourteen chapters in this book and we have wo... l License or stated otherwise. To view PDF files, there must be PDF Reader (Viewer) installed on your PC
- Subjective Mathematics 12th by Muhammad Shahbaz @fsc-part2-ptb
- , Punjab Textbook Board (PTB) Lahore, Pakistan.'' There are total seven (7) units in this book. These no
- FSc/ICS Part 2 Solutions @fsc
- r HSSC-II), Punjab Text Book Board Lahore.</lead> There are seven units in this book and we have work har
- Fluid Mechanics by Muhammad Usman Hamid @notes
- e and of the entire universe is in a fluid state. Therefore, it becomes essential for sciences and engine
- Unit 06: Permutation and Combination @math-11-nbf:sol
- on of $n$ different objects taken $r$ at a time. There are four exercises in this chapter. * **[[math
- Question 2 and 3, Review Exercise 6 @math-11-nbf:sol:unit06
- ===Question 3===== How many $3$-digit numbers are there which have $0$ at unit place? ** Solution. **
- Review Exercise (Solutions) @math-11-nbf:sol:unit06
- ] **Question 3.** How many $3$-digit numbers are there which has $0$ at unit place?\\ [[math-11-nbf:sol
- Exercise 6.3 (Solutions) @math-11-nbf:sol:unit06
- ways can 11 players be chosen out of 16 if\\ (i) there is no restriction. (ii) a particular player is al... -3-p7|Solution: Question 7 & 8]] **Question 8.** There are 10 points on a circle. Find the number of $(\... 2]] **Question 12.** For the post of 6 officers, there are 100 applicants, 2 posts are reserved for serving candidates and remaining for others.\\ Thereare 20 serving candidates among the applicants. In
- Question 11 and 12, Exercise 6.3 @math-11-nbf:sol:unit06
- ===Question 12===== For the post of $6$ officers, there are $100$ appliciants,\\ $2$ posts are reserved f... serving candidiates and remaining for others. \\ There are $20$ serving candidates among the appliciants
- Question 7 and 8, Exercise 6.3 @math-11-nbf:sol:unit06
- has exactly $2$ women. ** Solution. ** (i) If there are exactly $2$ women then there will be $3$ man in committee. Total possible ways $={ }^{4} C_{2} \tim... women. ** Solution. ** At least $2$ women means there could be more than $2$ women as well.\\ So we wil... having $2,3$ and four women in committee.\\ So if there are $2$ women we already had calculated $120$ pos
- Question 5 and 6, Exercise 6.3 @math-11-nbf:sol:unit06
- ny ways can $11$ players be chosen out of $16$ if there is no restriction. ** Solution. ** $11$ players... in ${ }^{16} C_{11}$ ways.\\ i.e. $4368$ ways are there to choose $11$ players.\\ =====Question 5(ii)===... ? ** Solution. ** Case I: If one man is chosen there will be two women in committee.\\ $1$ man may be ... \times{ }^{3} C_{2}=5 \times 3=15$\\ Case II: If there are $2$ men and one woman in committee. Total po
- Question 12 and 13, Exercise 6.2 @math-11-nbf:sol:unit06
- Possible arrangements of $7$ letters $=71=5040$\\ There are $3$ vowels $A, I$ and $E$ in word.\\ To find
- Question 16 and 17, Exercise 6.2 @math-11-nbf:sol:unit06
- rangements and\\ similarly for $5$ at units place there are $24$ arrangements.\\ Hence in total, we have
- Question 6(vi-ix), Exercise 6.1 @math-11-nbf:sol:unit06
- d{align*} So $33$ is the multiple of $2^{15}$ and therefore is divisible by $2^{15}$ =====Question 6(vii
- Topology: Handwritten Notes @notes
- -handwritten-notes.pdf}} </modal> </callout> ====There are more notes on the topology==== {{topic>Topolo