10 Most SAQ’s of Permutations and Combinations Chapter in Inter 2nd Year Maths-2A (TS/AP)

4 Marks

SAQ-1 : Simplify 34C5 +∑r=04(38-r)C4

$$G.E = \sum_{r=0}^{4} (38-r) C 4 + 34 C 5$$

$$G.E = [38 C 4 + 37 C 4 + 36 C 4 + 35 C 4 + 34 C 4] + 34 C 5$$

$$= [38 C 4 + 37 C 4 + 36 C 4 + 35 C 4] + [34 C 4 + 34 C 5]$$

$$= [38 C 4 + 37 C 4 + 36 C 4] + 35 C 4 + 35 C 5$$

$$= [38 C 4 + 37 C 4] + 36 C 4 + 36 C 5$$

$$= [38 C 4] + 37 C 4 + 37 C 5$$

$$= 38 C 4 + 38 C 5$$

$$= 39 C 5$$


SAQ-2 : Show that 25C4+∑r=04 (29-r)C3=30C4

$$G.E = \sum_{r=0}^{4} (29-r) C 3 + 25 C 4$$

$$G.E = [29 C 3 + 28 C 3 + 27 C 3 + 26 C 3 + 25 C 3] + 25 C 4$$

$$= [29 C 3 + 28 C 3 + 27 C 3 + 26 C 3] + [25 C 3 + 25 C 4]$$

$$= [29 C 3 + 28 C 3 + 27 C 3] + 26 C 3 + 26 C 4$$

$$= [29 C 3 + 28 C 3] + 27 C 3 + 27 C 4$$

$$= [29 C 3] + 28 C 3 + 28 C 4$$

$$= 29 C 3 + 29 C 4 = 30 C 4$$


SAQ-3 : Prove that 4nC2n/2nCn) =(1.3.5…..(4n-1)/[1.3.5….(2n-1)]2)

Given: $$L.H.S = \frac{4n C 2n}{2n C n} = \frac{4n!}{2n! \cdot 2n!} \cdot \frac{2n!}{n! \cdot n!} = \frac{(4n)!}{(2n!)^2} \times \frac{(n!)^2}{2n!}$$

This simplifies as follows:

$$= \frac{(4n)(4n – 1)(4n – 2)(4n – 3)(4n – 4)\ldots6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{[(2n)(2n – 1)(2n – 2)(2n – 3)\ldots4 \cdot 3 \cdot 2 \cdot 1]^2} = \frac{(n!)^2}{2n!}$$

$$= \frac{[(4n)(4n – 2)(4n – 4)\ldots(6)(4)(2)][(4n – 1)(4n – 3)\ldots5 \cdot 3 \cdot 1]}{[(2n)(2n – 2)\ldots4 \cdot 2]^2[(2n – 1)(2n – 3)\ldots(3)(1)]^2} \times \frac{(n!)^2}{2n!}$$

$$= \frac{[2^{2n}(2n)(2n – 1)(2n – 2)\ldots(3)(2)(1)][(4n – 1)(4n – 3)\ldots5 \cdot 3 \cdot 1]}{[2^n(n)(n – 1)\ldots(2)(1)]^2[(2n – 1)(2n – 3)\ldots(3)(1)]^2} \times \frac{(n!)^2}{2n!}$$

$$= \frac{[2^{2n}(2n!)]}{[2^{2n}(n!)^2]} \times \frac{[(4n – 1)(4n – 3)\ldots5 \cdot 3 \cdot 1]}{[(2n – 1)(2n – 3)\ldots(3)(1)]^2} \times \frac{(n!)^2}{2n!}$$

$$= 1 \times \frac{1.3.5\ldots(4n – 3)(4n – 1)}{[1.3.5\ldots(2n – 3)(2n – 1)]^2} = R.H.S$$


SAQ-4 : Find the number of ways of selecting a cricket team of 11 players from batsmen and 6 bowlers such that there will be atleast 5 bowlers in the team

A Team of 11 players with atleast 5 bowlers can be selected in the following compositions:

Bowlers(6)Batsmen(7)No. of selections
566 C 5 x 7 C 6 = 6 x 7 = 42
6 56 C 6 x 7 C 5 = 1 x 21 = 21
The total number of selections = 42 + 21 = 63

SAQ-5 : Find the number of ways of selecting 11 member cricket team from 7 batsmen, 6 bowlers and 2 wicket keepers so that the team contains 2 wicket keepers and atleast 4 bowlers

A team of 11 players with 2 wicket keepers and atleast 4 bowlers can be selected in the following compositions.

Keepers (2)Bowlers (6)Batsmen (7)No. of Selections
2452 C 2 x 6 C 4 x 7 C 5 = 1 x 15 x 21 = 315
2 542 C 2 x 6 C 5 x 7 C 4 = 1 x 6 x 35 = 210
2632 C 2 x 6 C 6 x 7 C 3 = 1 x 1 x 35 = 35
The total number of selections = 315 + 210 + 35 = 560

SAQ-6 : Find the number of ways of forming a committee of 5 members out of 6 Indians and 5 Americans so that always the Indians will be in majority in the committee

A 5 committee out of 6 Indians, 5 Americans with majority indians can be selected in the following compositions :

Indians (6)Americans (5)No. of Selections
506 C 5 x 5 C 0 = 6 x 1 = 6
416 C 4 x 5 C 1 = 15 x 5 = 75
326 C 3 x 5 C 2 = 20 x 10 = 200
The total number of selections = 6 + 75 + 200 = 281

SAQ-7 : If the letters of the word MASTER are permutted in all possible ways and the words thus formed are arranged in dictionary order, then find the rank of the word MASTER

The alphabetrical order of the letters of the word MASTER is

$$A, E, M, R, S, T$$

The number of words that begin with A —– = 5! = 120

The number of words that begin with E —– = 5! = 120

The number of words that begin with MAE — = 3! = 6

The number of words that begin with MAR — = 3! = 6

The number of words that begin with MASE — = 2! = 2

The number of words that begin with MASR — = 2! = 2

The next word is MASTEr = 1! = 1

Rank of the word MASTER

$$2(120) + 2(6) + 2(2) + 1 = 240 + 12 + 4 + 1 = 257$$


SAQ- 8 : If the letters of the word PRISON are permutted in all possible ways and the words thus formed are arranged in dictionary order, then find the rank of the word PRISON

The alphabetical order of the letters of the word PRISON is

$$I, N, O, P, R, S$$

The number of words that begin with I —– = 5! = 120

The number of words that begin with N —– = 5! = 120

The number of words that begin with O —– = 5! = 120

The number of words that begin with P I —- = 4! = 24

The number of words that begin with P N —- = 4! = 24

The number of words that begin with P O —- = 4! = 24

The number of words that begin with PRIN — = 2! = 2

The number of words that begin with PRIO — = 2! = 2

The number of words that begin with PRISN — = 1! = 1

The next word is PRISON = 1! = 1

Rank of the word PRISON = $$3(120) + 3(24) + 2(2) + 1 + 1 = 360 + 72 + 4 + 1 + 1 = 438$$


SAQ-9 : If the letters of the word thus formed are arranged in the dictionary order, find the rank of the word EAMCET

The alphabetical order of the letters of the word EAMCET is as follows :

$$A, C, E, E, M, T$$

The number of words that begin with A —– 5!/2! = 60

The number of words that begin with C —– 5!/2! = 60

The number of words that begin with E A C — 3! = 6

The number of words that begin with E A E — 3! = 6

The next word is EAMCET 1

Hence the rank of the word EAMCET is $$60 + 60 + 6 + 6 + 1 = 133$$


SAQ-10 : Find the sum of all 4 digit numbers that can be formed using digits 1,3,5,7,9

First we find the sum contributed by 9 in the total sum

The sum contributed by 9 when it is in units place is = 4 P 3 x 9

The sum contributed by 9 when it is in tens place is = 4 P 3 x 90

The sum contributed by 9 when it is in hundreds place is = 4 P 3 x 900

The sum contributed by 9 when it is in thousands place is = 4 P 3 x 9000

The sum contributed by 9 in the total sum is $$4P3 \times (9 + 90 + 900 + 9000) = 4P3 \times 9(1 + 10 + 100 + 1000) = 4P3 \times 9(1111)$$

Similarly, The sum contributed by 7 is $$4P3 \times 7(1111)$$

The sum contributed by 5 is $$4P3 \times 5(1111)$$

The sum contributed by 3 is $$4P3 \times 3(1111)$$

The sum contributed by 1 is $$4P3 \times 1(1111)$$

Hence from (1), (2), (3), (4), (5) the sum of all 4 digited numbers is

$$= 4P3 \times 1111(9 + 7 + 5 + 3 + 1)$$

$$= (4 \times 3 \times 2) \times 1111 \times 25$$

$$= 24 \times 25 \times 1111 = 666,600$$