Maharashtra Board Class 11 Maths Part 2 Chapter 6 Permutations and Combinations 6.6 Solutions

Question 1. Find the value of (i) \(^{15}C_4\)
Answer: Solution: \(^{15}C_4 = \frac{15!}{4!(15-4)!} = \frac{15!}{4!11!}\) \( = \frac{15 \times 14 \times 13 \times 12 \times 11!}{4 \times 3 \times 2 \times 1 \times 11!} \) \( = \frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} \) \( = 1365\)
(ii) \(^{80}C_2\)
Answer: Solution: \(^{80}C_2 = \frac{80!}{2!(80-2)!} = \frac{80!}{2!78!}\) \( = \frac{80 \times 79 \times 78!}{2 \times 78!} \) \( = 40 \times 79 = 3160\)
(iii) \(^{15}C_4 + ^{15}C_5\)
Answer: Solution: \(^{15}C_4 + ^{15}C_5 = ^{15}C_5 + ^{15}C_4 \) \( = ^{15}C_5 + ^{15}C_{5-1}\) \( = ^{16}C_5 \) [\( \because\) \(^nC_r + ^nC_{r-1} = ^{n+1}C_r\)] \( = 4368\)
(iv) \(^{20}C_{16} - ^{19}C_{16}\)
Answer: Solution: \(^{20}C_{16} - ^{19}C_{16} = \frac{20!}{16!4!} - \frac{19!}{16!3!}\) \( = \frac{20 \times 19!}{16! \times 4 \times 3!} - \frac{19!}{16!3!}\) \( = \frac{19!}{3!16!} \left[ \frac{20}{4} - 1 \right] \) \( = \frac{19!}{3!16!} (4)\) \( = \frac{19!}{3! (16)(15!)} \times 4 \) \( = \frac{19!}{4 (3!)(15!)}\) \( = \frac{19!}{4!15!} \) \( = ^{19}C_{15}\) or \(^{19}C_4 = 3876\) In simple words: This question demonstrates how to calculate combinations using the formula \(^nC_r = \frac{n!}{r!(n-r)!}\) and also applies the identity \(^nC_r + ^nC_{r-1} = ^{n+1}C_r\) for simplification.

🎯 Exam Tip: Remember the combination formula and key identities like Pascal's identity to efficiently solve such problems, as it simplifies calculations and saves time.

Question 2. Find n if (i) \(^6P_2 = n ^6C_2\)
Answer: Solution: \( \therefore ^6P_2 = n (^6C_2) \) \( \frac{6!}{(6-2)!} = n \frac{6!}{2!(6-2)!} \) \( \frac{6!}{4!} = n \frac{6!}{2!4!} \) \( \therefore n = 2! = 2 \times 1 = 2 \)
(ii) \(^{2n}C_3 : ^nC_2 = 52 : 3\)
Answer: Solution: \( \therefore ^{2n}C_3 : ^nC_2 = 52 : 3 \) \( \frac{(2n)!}{3!(2n-3)!} \div \frac{n!}{2!(n-2)!} = \frac{52}{3} \) \( \therefore \frac{(2n)!}{3!(2n-3)!} \times \frac{2!(n-2)!}{n!} = \frac{52}{3} \) \( \frac{(2n)(2n-1)(2n-2)(2n-3)!}{3 \times 2!(2n-3)!} \times \frac{2!(n-2)!}{n(n-1)(n-2)!} = \frac{52}{3} \) \( \frac{2n(2n-1).2(n-1)}{3} \times \frac{1}{n(n-1)} = \frac{52}{3} \) \( \therefore \frac{4(2n-1)}{3} = \frac{52}{3} \) \( \therefore 2n-1 = 13 \) \( 2n = 14 \) \( n = 7 \)
(iii) \(^nC_{n-3} = 84\)
Answer: Solution: \(^nC_{n-3} = 84\) \( \therefore \frac{n!}{(n-3)![n-(n-3)]!} = 84 \) \( \frac{n(n-1)(n-2)(n-3)!}{(n-3)! \times 3!} = 84 \) \( \therefore n(n-1)(n-2) = 84 \times 6 \) \( \therefore n(n-1)(n-2) = 9 \times 8 \times 7 \) Comparing on both sides, we get \( \therefore n = 9 \) In simple words: This question involves solving for 'n' using both permutations and combinations formulas and their relationships. Part (iii) shows how to find 'n' by setting up a combination equation and factoring.

🎯 Exam Tip: When solving for 'n' in combination/permutation equations, always simplify expressions fully and compare factorials or products of consecutive integers to find the value of 'n' efficiently.

Question 3. Find r if \(^{14}C_{2r} : ^{10}C_{2r-4} = 143 : 10\)
Answer: Solution: \( \therefore ^{14}C_{2r} : ^{10}C_{2r-4} = 143 : 10 \) \( \frac{14!}{2r!(14-2r)!} \div \frac{10!}{(2r-4)!(10-(2r-4))!} = \frac{143}{10} \) \( \therefore \frac{14!}{2r!(14-2r)!} \times \frac{(2r-4)!(14-2r)!}{10!} = \frac{143}{10} \) \( \therefore \frac{14 \times 13 \times 12 \times 11 \times 10!}{2r(2r-1)(2r-2)(2r-3)(2r-4)!(14-2r)!} \times \frac{(2r-4)!(14-2r)!}{10!} = \frac{143}{10} \) \( \frac{14 \times 13 \times 12 \times 11}{2r(2r-1)(2r-2)(2r-3)} = \frac{143}{10} \) \( \therefore 2r(2r-1)(2r-2)(2r-3) = 14 \times 12 \times 10 \) \( \therefore 2r(2r-1)(2r-2)(2r-3) = 8 \times 7 \times 6 \times 5 \) Comparing on both sides, we get \( \therefore r = 4 \) In simple words: This problem asks us to find the value of 'r' given a ratio of two combination expressions. We expand the combinations using their formula, simplify by canceling terms, and then compare the resulting products to solve for 'r'.

🎯 Exam Tip: When dealing with ratios of combinations, express them as a fraction and simplify before cross-multiplication. Look for consecutive integer products to easily compare and find the unknown variable.

Question 4. Find n and r if. (i) \(^nP_r = 720\) and \(^nC_{n-r} = 120\)
Answer: Solution: \(^nP_r = 720\) \( \therefore \frac{n!}{(n-r)!} = 720 \) ...(i) Also, \(^nC_{n-r} = 120\) \( \therefore \frac{n!}{(n-r)![n-(n-r)]!} = 120 \) \( \frac{n!}{r!(n-r)!} = 120 \) ...(ii) Dividing (i) by (ii), we get \( \frac{\frac{n!}{(n-r)!}}{\frac{n!}{r!(n-r)!}} = \frac{720}{120} \) \( r! = 6 \) \( \therefore r = 3 \) Substituting \( r = 3 \) in (i), we get \( \frac{n!}{(n-3)!} = 720 \) \( \frac{n(n-1)(n-2)(n-3)!}{(n-3)!} = 720 \) \( n(n-1)(n-2) = 10 \times 9 \times 8 \) \( \therefore n = 10 \)
(ii) \(^nC_{r-1} : ^nC_r : ^nC_{r+1} = 20 : 35 : 42\)
Answer: Solution: \( \therefore ^nC_{r-1} : ^nC_r : ^nC_{r+1} = 20 : 35 : 42 \) \( \frac{^nC_{r-1}}{^nC_r} = \frac{20}{35} = \frac{4}{7} \) \( \therefore \frac{\frac{n!}{(r-1)![n-(r-1)]!}}{\frac{n!}{r!(n-r)!}} = \frac{4}{7} \) \( \therefore \frac{n!}{(r-1)!(n+1-r)!} \times \frac{r!(n-r)!}{n!} = \frac{4}{7} \) \( \frac{r(r-1)!(n-r)!}{(r-1)!(n+1-r)(n-r)!} = \frac{4}{7} \) \( \frac{r}{n-r+1} = \frac{4}{7} \) \( \therefore 7r = 4(n+1-r) \) \( 7r = 4n+4-4r \) \( 11r = 4n+4 \) ...(i) Also, \( \frac{^nC_r}{^nC_{r+1}} = \frac{35}{42} = \frac{5}{6} \) \( \therefore \frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r+1)![n-(r+1)]!}} = \frac{5}{6} \) \( \therefore \frac{n!}{r!(n-r)!} \times \frac{(r+1)!(n-r-1)!}{n!} = \frac{5}{6} \) \( \frac{(r+1)r!(n-r-1)!}{r!(n-r)(n-r-1)!} = \frac{5}{6} \) \( \frac{r+1}{n-r} = \frac{5}{6} \) \( \therefore 6(r+1) = 5(n-r) \) \( 6r+6 = 5n-5r \) \( 11r = 5n-6 \) ...(ii) From (i), \( 4n+4 = 11r \) From (ii), \( 5n-6 = 11r \) So, \( 4n+4 = 5n-6 \) \( 4+6 = 5n-4n \) \( \therefore n = 10 \) Substituting \( n = 10 \) in (i), \( 11r = 4(10)+4 \) \( 11r = 44 \) \( \therefore r = 4 \) In simple words: This problem requires finding both 'n' and 'r' by using given permutation and combination values. We use the relationship between permutations and combinations to find 'r' first, then substitute it back into one of the original equations to find 'n'. Part (ii) involves setting up a system of equations from given ratios of consecutive combinations and solving them simultaneously.

🎯 Exam Tip: For problems involving ratios of consecutive combinations, utilize the identity \( \frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r} \) to simplify expressions quickly and set up linear equations for 'n' and 'r'.

Question 5. If \(^nP_r = 1814400\) and \(^nC_r = 45\), find r.
Answer: Solution: Given, \(^nP_r = 1814400\) and \(^nC_r = 45\) We know that \(^nP_r = r! \times ^nC_r\) \( \therefore r! = \frac{^nP_r}{^nC_r} \) \( r! = \frac{1814400}{45} \) \( r! = 40320 \) \( \therefore r! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \) \( \therefore r! = 8! \) \( \therefore r = 8 \) In simple words: This question uses the fundamental relationship between permutations and combinations, \(^nP_r = r! \times ^nC_r\), to find 'r'. By dividing the given permutation value by the combination value, we find 'r!', which then allows us to determine 'r'.

🎯 Exam Tip: Always remember the relation \(^nP_r = r! \times ^nC_r\). It's a powerful tool to solve problems where both permutations and combinations are given, making it easy to find 'r' (or 'n' if other values are known).

Question 6. If \(^nC_{r-1} = 6435\), \(^nC_r = 5005\), \(^nC_{r+1} = 3003\), find \(^nC_5\).
Answer: Solution: Given, \(^nC_{r-1} = 6435\), \(^nC_r = 5005\), \(^nC_{r+1} = 3003\) \( \therefore \frac{^nC_{r-1}}{^nC_r} = \frac{6435}{5005} = \frac{9}{7} \) \( \therefore \frac{\frac{n!}{(r-1)![n-(r-1)]!}}{\frac{n!}{r!(n-r)!}} = \frac{9}{7} \) \( \therefore \frac{n!}{(r-1)!(n-r+1)!} \times \frac{r!(n-r)!}{n!} = \frac{9}{7} \) \( \frac{r(r-1)!(n-r)!}{(r-1)!(n-r+1)(n-r)!} = \frac{9}{7} \) \( \frac{r}{n-r+1} = \frac{9}{7} \) \( 7r = 9(n-r+1) \) \( 7r = 9n-9r+9 \) \( 16r-9n = 9 \) ...(i) Now, \( \frac{^nC_r}{^nC_{r+1}} = \frac{5005}{3003} = \frac{5}{3} \) \( \therefore \frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r+1)![n-(r+1)]!}} = \frac{5}{3} \) \( \therefore \frac{n!}{r!(n-r)!} \times \frac{(r+1)!(n-r-1)!}{n!} = \frac{5}{3} \) \( \frac{(r+1)r!(n-r-1)!}{r!(n-r)(n-r-1)!} = \frac{5}{3} \) \( \frac{r+1}{n-r} = \frac{5}{3} \) \( 3(r+1) = 5(n-r) \) \( 3r+3 = 5n-5r \) \( 8r-5n = -3 \) ...(ii) Solving (i) and (ii): From (i): \( 9n = 16r-9 \) From (ii): \( 5n = 8r+3 \) Multiply (ii) by 9 and (i) by 5: \( 45n = 72r+27 \) \( 45n = 80r-45 \) So, \( 72r+27 = 80r-45 \) \( 27+45 = 80r-72r \) \( 72 = 8r \) \( \therefore r = 9 \) Substitute \( r = 9 \) in (ii): \( 8(9)-5n = -3 \) \( 72-5n = -3 \) \( 75 = 5n \) \( \therefore n = 15 \) Now we need to find \(^nC_5 = ^{15}C_5\) \( ^{15}C_5 = \frac{15!}{5!(15-5)!} = \frac{15!}{5!10!} \) \( = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10!}{5 \times 4 \times 3 \times 2 \times 1 \times 10!} \) \( = 3 \times 7 \times 13 \times 11 \) \( = 3003 \) (Wait, there's a small mistake in the OCR's calculation for 9C5, I will follow the OCR's provided calculation, but the prompt asks for `^nC5` and gives `9C5`. Since `n=15`, it should be `15C5`. I will follow the OCR exactly if the calculation is incorrect, as per VERBATIM rule.) Let me re-check the OCR for the final calculation: OCR has: `^nC5 = 9C5 = 9! / (4!5!) = 9x8x7x6x5! / (4x3x2x1x5!) = 126` This means the OCR used n=9, not n=15. But n was calculated as 15. This is a discrepancy. Given the rule "Extract every word exactly as written", I should transcribe the OCR's calculation, even if it has a logical inconsistency regarding the derived 'n'. It appears the OCR text snippet for `^nC5` calculation directly uses `9C5` without using the `n=15` derived. I will output what the OCR shows. \(^nC_5 = ^9C_5 = \frac{9!}{4!5!}\) \( = \frac{9 \times 8 \times 7 \times 6 \times 5!}{4 \times 3 \times 2 \times 1 \times 5!} \) \( = 126 \) In simple words: This problem involves finding 'n' and 'r' from a series of consecutive combination values by setting up two ratios. These ratios lead to two linear equations in 'n' and 'r', which are then solved simultaneously. Finally, the value of \(^nC_5\) is calculated using the determined 'n' (even if the OCR has a discrepancy in the final calculation step, it is reproduced as found).

🎯 Exam Tip: When given multiple consecutive combination values, the ratio method \(\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}\) is crucial. Systematically solve the simultaneous equations for 'n' and 'r' to ensure accuracy, and always double-check the final calculation step.

Question 7. Find the number of ways of drawing 9 balls from a bag that has 6 red balls, 5 green balls and 7 blue balls so that 3 balls of every colour are drawn.
Answer: Solution: 9 balls are to be selected from 6 red, 5 green, 7 blue balls such that the selection consists of 3 balls of each colour. \( \therefore \) 3 red balls can be selected from 6 red balls in \(^6C_3\) ways. 3 green balls can be selected from 5 green balls in \(^5C_3\) ways. 3 blue balls can be selected from 7 blue balls in \(^7C_3\) ways. \( \therefore \) Number of ways selection can be done if the selection consists of 3 balls of each colour \( = ^6C_3 \times ^5C_3 \times ^7C_3 \) \( = \frac{6!}{3!3!} \times \frac{5!}{3!2!} \times \frac{7!}{3!4!} \) \( = \frac{6 \times 5 \times 4 \times 3!}{3 \times 2 \times 1 \times 3!} \times \frac{5 \times 4 \times 3!}{3 \times 2 \times 1 \times 2!} \times \frac{7 \times 6 \times 5 \times 4!}{3 \times 2 \times 1 \times 4!} \) \( = 20 \times 10 \times 35 \) \( = 7000 \) In simple words: To find the number of ways to select 9 balls with 3 of each color from a given set, we calculate the combinations for each color separately and then multiply these results together.

🎯 Exam Tip: When selecting items with specific conditions from multiple groups, use the "multiplication principle." Calculate combinations for each group independently, then multiply the results to get the total number of ways.

Question 8. Find the number of ways of selecting a team of 3 boys and 2 girls from 6 boys and 4 girls.
Answer: Solution: There are 6 boys and 4 girls. A team of 3 boys and 2 girls is to be selected. \( \therefore \) 3 boys can be selected from 6 boys in \(^6C_3\) ways. 2 girls can be selected from 4 girls in \(^4C_2\) ways. \( \therefore \) Number of ways the team can be selected \( = ^6C_3 \times ^4C_2 \) \( = \frac{6!}{3!3!} \times \frac{4!}{2!2!} \) \( = \frac{6 \times 5 \times 4 \times 3!}{3 \times 2 \times 1 \times 3!} \times \frac{4 \times 3 \times 2!}{2 \times 1 \times 2!} \) \( = 20 \times 6 \) \( = 120 \) \( \therefore \) The team of 3 boys and 2 girls can be selected in 120 ways. In simple words: To select a team with a specific number of boys and girls from distinct groups, calculate the number of ways to choose boys and girls separately using combinations, then multiply these results.

🎯 Exam Tip: For problems involving selections from different categories (e.g., boys and girls), calculate the combinations for each category independently and then multiply the results to find the total number of ways to form the desired group.

Question 9. After a meeting, every participant shakes hands with every other participants. If the number of handshakes is 66, find the number of participants in the meeting.
Answer: Solution: Let there be n participants present in the meeting. A handshake occurs between 2 persons. \( \therefore \) Number of handshakes \( = ^nC_2 \) Given 66 handshakes were exchanged. \( \therefore 66 = ^nC_2 \) \( \therefore 66 = \frac{n!}{2!(n-2)!} \) \( \therefore 66 \times 2 = \frac{n(n-1)(n-2)!}{(n-2)!} \) \( \therefore 132 = n(n-1) \) \( \therefore n(n-1) = 12 \times 11 \) Comparing on both sides, we get \( n = 12 \) \( \therefore \) 12 participants were present at the meeting. In simple words: This problem relates the number of handshakes to the number of participants using the combination formula \(^nC_2\). By setting the formula equal to the given number of handshakes, we solve for 'n' by finding two consecutive integers whose product matches the result.

🎯 Exam Tip: The "handshake problem" is a classic combination application. Remember that if 'n' people are present, the number of handshakes is \(^nC_2\). Formulate the equation and look for consecutive integers to solve for 'n'.

Question 10. If 20 points are marked on a circle, how many chords can be drawn?
Answer: Solution: To draw a chord we need to join two points on the circle. There are 20 points on a circle. \( \therefore \) Total number of chords possible from these points \( = ^{20}C_2 \) \( = \frac{20!}{2!18!} \) \( = \frac{20 \times 19 \times 18!}{2 \times 1 \times 18!} \) \( = 190 \) In simple words: Since a chord is formed by connecting any two distinct points on a circle, we use the combination formula to select 2 points out of the total 20 points available.

🎯 Exam Tip: Drawing a chord always requires selecting exactly two points. So, for 'n' points, the number of chords is always \(^nC_2\). This is a direct application of the combination formula.

Question 11. Find the number of diagonals of an n-sided polygon. In particular, find the number of diagonals when (i) n = 10 (ii) n = 15 (iii) n = 12
Answer: Solution: In n-sided polygon, there are 'n' points and 'n' sides. . \( \therefore \) Through 'n' points we can draw \(^nC_2\) lines including sides. \( \therefore \) Number of diagonals in n sided polygon \( = ^nC_2 - n \) (\( \therefore \) n = number of sides) (i) n = 10, \(^nC_2 - n = ^{10}C_2 - 10 \) \( = \frac{10 \times 9}{1 \times 2} - 10 \) \( = 45 - 10 = 35 \)
(ii) n = 15, \(^nC_2 - n = ^{15}C_2 - 15 \) \( = \frac{15 \times 14}{1 \times 2} - 15 \) \( = 105 - 15 = 90 \)
(iii) n = 12, \(^nC_2 - n = ^{12}C_2 - 12 \) \( = \frac{12 \times 11}{1 \times 2} - 12 \) \( = 66 - 12 = 54 \) In simple words: The total number of lines that can be drawn by connecting 'n' vertices of a polygon is \(^nC_2\). Since 'n' of these lines are the sides of the polygon, the number of diagonals is found by subtracting 'n' from \(^nC_2\).

🎯 Exam Tip: The formula for the number of diagonals in an n-sided polygon is \( \frac{n(n-3)}{2} \) or \(^nC_2 - n\). Ensure you subtract the number of sides (n) from the total possible lines to get only the diagonals.

Question 12. There are 20 straight lines in a plane so that no two lines are parallel and no three lines are concurrent. Determine the number of points of intersection.
Answer: Solution: There are 20 lines such that no two of them are parallel and no three of them are concurrent. Since no two lines are parallel \( \therefore \) they intersect at a point \( \therefore \) Number of points of intersection if no two lines are parallel and no three lines are concurrent \( = ^{20}C_2 \) \( = \frac{20!}{2!18!} \) \( = \frac{20 \times 19 \times 18!}{2 \times 1 \times 18!} \) \( = 190 \) In simple words: When no two lines are parallel and no three lines intersect at the same point, each pair of lines creates a unique intersection point. Therefore, the number of intersection points is found by choosing 2 lines out of the total 20 lines.

🎯 Exam Tip: For 'n' lines in a plane, if no two are parallel and no three are concurrent, the number of intersection points is \(^nC_2\). This is a standard combination problem where each unique pair of lines forms one intersection.

Question 13. Ten points are plotted on a plane. Find the number of straight lines obtained by joining these points if (i) no three points are collinear (ii) four points are collinear
Answer: Solution: There are 10 points on a plane. (i) No three of them are collinear: Since a line is obtained by joining 2 points, number of lines passing through these points if no three points are collinear \( = ^{10}C_2 \) \( = \frac{10!}{2!8!} \) \( = \frac{10 \times 9 \times 8!}{2 \times 1 \times 8!} \) \( = 5 \times 9 \) \( = 45 \)
(ii) When 4 of them are collinear: \( \therefore \) Number of lines passing through these points if 4 points are collinear \( = ^{10}C_2 - ^4C_2 + 1 \) \( = 45 - \frac{4!}{2!2!} + 1 \) \( = 45 - \frac{4 \times 3 \times 2!}{2 \times 1 \times 2!} + 1 \) \( = 45 - 6 + 1 \) \( = 40 \) In simple words: For part (i), if no three points are collinear, any two points form a unique line, so we use \(^{10}C_2\). For part (ii), when some points are collinear, we subtract the lines formed by those collinear points (\(^4C_2\)) from the total possible lines, and then add 1 back because the collinear points themselves form one single line.

🎯 Exam Tip: When points are collinear, the general formula \(^nC_2\) must be adjusted. Subtract \(^kC_2\) for every 'k' collinear points, and add 1 back for the single line they form. This ensures accurate line counting in such configurations.

Question 14. Find the number of triangles formed by joining 12 points if (i) no three points are collinear (ii) four points are collinear
Answer: Solution: There are 12 points on the plane (i) When no three of them are collinear: Since a triangle can be drawn by joining any three non-collinear points. \( \therefore \) Number of triangles that can be obtained from these points \( = ^{12}C_3 \) \( = \frac{12!}{3!9!} \) \( = \frac{12 \times 11 \times 10 \times 9!}{3 \times 2 \times 1 \times 9!} \) \( = 220 \)
(ii) When 4 of these points are collinear: \( \therefore \) Number of triangles that can be obtained from these points \( = ^{12}C_3 - ^4C_3 \) \( = 220 - \frac{4!}{3!1!} \) \( = 220 - \frac{4 \times 3!}{3! \times 1} \) \( = 220 - 4 \) \( = 216 \) In simple words: To form a triangle, three non-collinear points are needed. In part (i), we simply select 3 points out of 12 using combinations. In part (ii), if some points are collinear, those collinear points cannot form a triangle, so we subtract the combinations of choosing 3 points from the collinear group from the total possible triangles.

🎯 Exam Tip: To count triangles, use \(^nC_3\). If 'k' points are collinear, subtract \(^kC_3\) from the total \(^nC_3\) because collinear points cannot form a triangle. This is a common adjustment for geometric combination problems.

Question 15. A word has 8 consonants and 3 vowels. How many distinct words can be formed if 4 consonants and 2 vowels are chosen?
Answer: Solution: Out of 8 consonants, 4 can be selected in \(^8C_4\) \( = \frac{8!}{4!4!} \) \( = \frac{8 \times 7 \times 6 \times 5 \times 4!}{4 \times 3 \times 2 \times 1 \times 4!} \) \( = 70 \) ways From 3 vowels, 2 can be selected in \(^3C_2\) \( = \frac{3!}{2!1!} \) \( = \frac{3 \times 2!}{2! \times 1} \) \( = 3 \) ways Now, to form a word, these 6 letters (i.e., 4 consonants and 2 vowels) can be arranged in \(^6P_6\) i.e., 6! ways. \( \therefore \) Total number of words that can be formed \( = 70 \times 3 \times 6! \) \( = 70 \times 3 \times 720 \) \( = 151200 \) \( \therefore \) 151200 words of 4 consonants and 2 vowels can be formed. In simple words: First, we select the required number of consonants and vowels using combinations. Then, since these chosen letters form a "word", their arrangement matters, so we find the permutations of the selected letters. The final answer is the product of the selections and the arrangement.

🎯 Exam Tip: For problems involving forming words from a pool of letters with categories (like consonants and vowels), first calculate the number of ways to *select* the required items (combinations), and then calculate the number of ways to *arrange* them (permutations or factorials) to find the total distinct words.