IPMAT - Indore questions on Modern Math - Permutation & Combination

Let there be n players in each of these 4 groups.
∴ There are total 4n players.

Let the groups be called G1, G2, G3 & G4.

Now, a player from G1, say A, plays with other (n - 1) players from G1 exactly once.
Hence, he plays (n -1) games.

A will play with each of n players of G2 twice, hence total 2n games.

A will play with each of n players of G3 thrice, hence total 3n games.

A will play with each of n players of G4 four times, hence total 4n games.

∴ A will play a total of (n - 1) + 2n + 3n + 4n = 10n - 1 games.

Now, each of the 4n players will play (10n - 1) games.
∴ There will be total 4n × (10n - 1)/2 games [/2 because each game is counted twice]
⇒ There are total 2n(10n - 1) games

⇒ 2n(10n - 1) > 1000
⇒ n(10n - 1) > 500

The least possible value of n satisfying the above inequality is 8.

Hence, 8.

We know (x + y)ⁿ = ⁿC₀ × xⁿ + ⁿC₁ × x^n-1 × y¹ + ⁿC₂ × x^n-2 × y² + ... + ⁿC_n × yⁿ

Let the three consecutive coefficients be ⁿC_r-1, ⁿC_r & ⁿC_r+1

Now, ⁿC_r-1 : ⁿC_r = 1 : 9
⇒ $\frac{\mathrm{n}!}{(\mathrm{r}-1)!\times (\mathrm{n}-\mathrm{r}+1)!}$ : $\frac{\mathrm{n}!}{\left(\mathrm{r}\right)!\times (\mathrm{n}-\mathrm{r})!}$ = 1 : 9

⇒ $\frac{\mathrm{r}}{\mathrm{n}-\mathrm{r}+1}$ = $\frac{1}{9}$

⇒ 9r = n - r + 1

⇒ 10r = n + 1   ...(1)

Also, ⁿC_r : ⁿC_r+1 = 9 : 63 = 1 : 7
⇒ $\frac{\mathrm{n}!}{\left(\mathrm{r}\right)!\times (\mathrm{n}-\mathrm{r})!}$ : $\frac{\mathrm{n}!}{(\mathrm{r}+1)!\times (\mathrm{n}-\mathrm{r}-1)!}$ = 1 : 7

⇒ $\frac{\mathrm{r}+1}{\mathrm{n}-\mathrm{r}}$ = $\frac{1}{7}$

⇒ 7r + 7 = n - r

⇒ 8r = n - 7   ...(2)

Solving (1) & (2), we get

n = 39 & r = 4

Hence, 39.

7 points lie on a line and 5 points on another parallel line.

A triangle can be formed by 2 points on one of these lines and third point on the other line.

Case 1: 2 points are chose on line with 7 points and 3rd point is chosen on line with 5 points.
⇒ Number of such triangles = ⁷C₂ × ⁵C₁ = 21 × 5 = 105

Case 2: 1 point is chose on line with 7 points and 2 points are chosen on line with 5 points.
⇒ Number of such triangles = ⁷C₁ × ⁵C₂ = 7 × 10 = 70

∴ Total number of traingles = 105 + 70 = 175.

Hence, 175.

Sum of the coefficients of all the terms in the expansion of (ax − b)ⁿ can be obtained by substituting x = 1.

∴ Sum of the coefficients of all the terms in the expansion of (5x − 9)⁴ = (5 - 9)⁴ = 4⁴ = 256.

Hence, 256.

Let us form a group X of Mr. and Mr. Ahuja. They can be arranged in 2! = 2 ways.

Now, X and 4 other persons can be seated around a circle in 4! = 24 ways.

Now we have total 5 people (X is counted as 1 only, no one should sit between Ahujaas) and there are 5 places between these 5 people for Mr. and Mrs. Sharma to sit.

We need to select 2 of these 5 places for Mr. and Mrs. Sharma. This can be done in ⁵C₂ = 10 ways.
Mr. and Mrs. Sharam can sit in these 2 places in 2! = 2 ways.

∴ Total number of ways = 2 × 24 × 10 × 2 = 960.

Hence, 960.

Vowels: A, A, E, E
Consonants = M, N, G, M, N, T

We first arrange the consonants in $\frac{6!}{2!\times 2!}$ = 180 ways.

This can be represented as: C C C C C C

Now we have 6 places for 4 vowels:  | C | C | C | C | C | C |

Out of 7 places we can select 4 places for vowels in 7C4 = 35 ways

4 vowels can be arranged in these 4 places in $\frac{4!}{2!\times 2!}$ = 6 ways.

∴ Total number of ways of arranging = 180 × 35 × 8 = 37,800

Hence, option (c).

There are 5 parallel lines and there are another 'n' lines which are parallel but perpendicular to the first 5 lines.

∴ Each of the first 5 lines (say they are all horizontal lines) are perpendicular to each of the other n lines (say they are vertical lines).

To form a rectangle we need two horizontal and two vertical lines.

∴ Number of rectanges = (number of ways of selecting 2 horizontal lines) × (number of ways of selecting 2 vertical lines) 

⇒ 360 = ⁵C₂ × ⁿC₂ 

⇒ 360 = 10 × n(n - 1)/2 

⇒ 72 = n(n - 1)

⇒ n = 9

Hence, 9.

Total number of lines that can be formed using 10 points = ¹⁰C₂ = 45 lines.

But out of these 10, 5 points are collinear and hence will give only 1 line instead of ¹⁰C₂ = 10.

∴ Number of unique lines = 45 - 10 + 1 = 36.

Hence, option (d).

Case 1: All 4 objects are distinct.
Number of ways of choosing 4 distinct objects out of 9 = ⁹C₄.
∴ Total number of such ways = 1 way.

Case 2: 3 objects are distinct and 1 is indistinguishable.
Number of ways of choosing 3 distinct objects out of 9 = ⁹C₃.
Number of ways of choosing 1 indistinguishable object out of 4 = 1.
∴ Total number of such ways = ⁹C₃ × 1 = ⁹C₃.

Case 3: 2 objects are distinct and 2 are indistinguishable.
Number of ways of choosing 2 distinct objects out of 9 = ⁹C₂.
Number of ways of choosing 2 indistinguishable object outs of 4 = 1.
∴ Total number of such ways = ⁹C₂ × 1 = ⁹C₂.

Case 4: 1 object is distinct and 3 are indistinguishable.
Number of ways of choosing 1 distinct objects out of 9 = ⁹C₁.
Number of ways of choosing 3 indistinguishable objects out of 4 = 1.
∴ Total number of such ways = ⁹C₁ × 1 = ⁹C₁.

Case 5: All 4 are indistinguishable.
Number of ways of choosing 0 distinct objects out of 9 = ⁹C₀.
Number of ways of choosing 4 indistinguishable objects out of 4 = 1.
∴ Total number of such ways = ⁹C₁ × 1 = ⁹C₁.

⇒ Total number of ways = ⁹C₄ + ⁹C₃ + ⁹C₂ + ⁹C₁ + ⁹C₀ = 256

Hence, 256.