# 65. Permutations and Combinations Solutions Part 6¶

1. L.H.S. $$= 2.6.10.\ldots(4n - 6)(4n - 2)$$

$$= 2^n(1.3.5.\ldots.(2n - 3)(2n - 1))$$

$$= \frac{2^n(1.2.3.4.5.\ldots(2n - 3)(2n - 2)(2n - 1)2n)}{2.4.6.\ldots.2n}$$

$$= \frac{2^n(1.2.3.4.5.\ldots(2n - 3)(2n - 2)(2n - 1)2n)}{2^n(1.2.3.\ldots.n)}$$

$$= \frac{2n!}{n!} =$$ R.H.S.

2. L.H.S. $$= {}^{47}C_4 + \sum_{i=0}^3{}^{50- i}C_3 + \sum_{j=1}^5{}^{56 - j}C_{53 - j}$$

$$= {}^{47}C_4 + {}^{47}C_3 + \sum_{i=0}^2{}^{50 - i}C_3 + \sum_{j=1}^5{}^{56 - j}C_{53 - j}$$

[ We know that $${}^nC_r + {}^nC_{r - 1} = {}^{n + 1}C_r$$ ]

$$= {}^{48}C_4 + \sum_{i=0}^2{}^{50- i}C_3 + \sum_{j=1}^5{}^{56 - j}C_{53 - j}$$

$$= {}^{48}C_4 + {}^{48}C_3 + \sum_{i=0}^1{}^{50 - i}C_3 + \sum_{j=1}^5{}^{56 - j}C_{53 - j}$$

$$= {}^{49}C_4 + \sum_{i=0}^1{}^{50- i}C_3 + \sum_{j=1}^5{}^{56 - j}C_{53 - j}$$

Following similarly we obtain,

$$= {}^{51}C_4 + \sum_{j=1}^5{}^{56 - j}C_{53 - j}$$

[ We know that $${}^nC_r = {}^nC_{n - r}$$ ]

Thus, we can rewrite the previous expression as

$$= {}^{51}C_4 + \sum_{j=1}^5{}^{56 - j}C_3$$

Again, we follow as previously to obtain

$$= {}^{57}C_4$$

3. L.H.S. $$= {}^nC_k + \sum_{j=0}^m{}^{n+j}C_{k - 1}$$

[ We know that $${}^nC_r + {}^nC_{r - 1} = {}^{n + 1}C_r$$ ]

Rewriting given expression,

$$= {}^nC_k + {}^nC_{k - 1} + \sum_{j=1}^m{}^{n+j}C_{k - 1}$$

$$= {}^{n + 1}C_k + \sum_{j=1}^m{}^{n+j}C_{k - 1}$$

Repeating,

$$= {}^{n + 1}C_k + {}^{n + 1}C_{k - 1} + \sum_{j=2}^m{}^{n+j}C_{k - 1}$$

$$= {}^{n + 2}C_k + \sum_{j=2}^m{}^{n+j}C_{k - 1}$$

Repeating this for remaining values of $$j$$, we obtain

$$= {}^{n + m + 1}C_k$$

4. L.H.S. $$= {}^mC_1 +{}^{m+1}C_2 + \ldots +{}^{m + n - 1}C_n$$

$$= {}^mC_0 + {}^mC_1 +{}^{m+1}C_2 + \ldots +{}^{m + n - 1}C_n - {}^mC_0$$

[ We know that $${}^nC_r + {}^nC_{r - 1} = {}^{n + 1}C_r$$ and $${}^nC_0 =1$$ ]

Applying the above formula, we get

$$= {}^{m+1}C_1 + {}^{m+1}C_2 + \ldots +{}^{m + n - 1}C_n - 1$$

$$= {}^{m+2}C_2 + {}^{m+2}C_3 + \ldots +{}^{m + n - 1}C_n - 1$$

Repeating this we obain,

$$= {}^{m + n}C_n - 1$$

We obtain the same value by computing R.H.S. in similar manner which turns out to be,

$$= {}^{m + n}C_m - 1$$

$$\because {}^nC_r = {}^nC_{n - r}$$

L.H.S. = R.H.S.

5. For a number to be divisible by $$25$$ last two digits should be one of $$00, 25, 50, 75$$.

Since we have only one $$0$$, last two digits cannot be $$00$$.

Case I: When last two digits are $$25$$

No. of ways to fill ten thousand’s place $$= 5$$ because we cannot use $$0$$

No. of ways to fill thousand’s place $$= 5$$

No. of ways to fill hundred’s place $$= 4$$

$$\therefore$$ Total no. of ways $$= 100$$

Case II: When last two digits are $$75$$

Like Case I total no. of such numbers $$= 100$$

Case III: When last two digits are $$50$$

No. of ways to fill ten thousand’s place $$= 6$$

No. of ways to fill thousand’s place $$= 5$$

No. of ways to fill hundred’s place $$= 4$$

Total no. of such numbers $$= 120$$

$$\therefore$$ Desired answer is $$320$$

6. For a number to be divisible by $$4$$ last two digits should be $$12, 24, 32, 52$$

No. of ways to fill ten thousand’s place $$= 3$$

No. of ways to fill thousand’s place $$= 2$$

No. of ways to fill hundred’s place $$= 1$$

$$\therefore$$ Desired answer $$= 4*3*2 = 24$$

7. For a number to be divisible by $$3$$ sum of digits should be divisible by $$3$$. The combination of such digits are $$123, 234, 345, 135, 1245$$

Considering $$123$$ with $$0$$

No. of ways to fill thousand’s place $$= 3$$

No. of ways to fill hundred’s place $$= 3$$

No. of ways to fill ten’s place $$= 2$$

$$\therefore$$ total no. of permutations $$= 18$$

Total no. of numbers including $$0 = 4*18 = 72$$

Total no. of permutation of $$1245 = 4! = 24$$

Total no. of numbers divisible by $$6 = 24 + 72 = 96$$

For these numbers to be divisible by $$6$$ these should be even.

Case I: When out three digit no. has two odd digits

If $$0$$ is at unit’s place, total no. of such numbers $$= 3 * 2$$

If $$0$$ is not at unit’s place, total no. of such numbers $$= 2 * 2$$

There are three such numbers(with two odd digits), therefore total no. of numbers $$= 3 * (3 * 2 + 2 *2) = 30$$

Case II: When there are two even numbers in our combination of three digit numbers

If $$0$$ is at unit’s place, total no. of such numbers $$= 3 * 2$$

If $$0$$ is not at unit’s place, total no. of such numbers $$= 2 * 2$$

There are one such numbers(with one odd digits), therefore total no. of numbers $$= 1 * (3 * 2 + 2 *2) = 10$$

Case III: When the no. is permutation of $$1245$$

Half of these will have even no. at unit’s place i.e. $$12$$

Therefore, total no. of numbers divisible by $$6 = 40 + 12 = 52$$

8. If the two $$3$$ were different then,

Unit’s place can be filled in $$3$$ ways.

Thousand’s place can be filled in $$3$$ ways.

Hundred’s place can be filled in $$2$$ ways.

Therefore, total no. of $$4$$ digit numbers $$= 3 * 3 *2 = 18$$

However, the two $$3$$ are same, therefore total no. of numbers $$= \frac{18}{2} = 9$$

Out of $$9$$ numbers at unit’s, ten’s and hundred’s place there will be $$3$$ will come four times, $$1$$ will come twice and $$0$$ will come thrice.

On thousand’s place $$3$$ will come six times and $$1$$ will come three times.

Sum of digits at unit’s place $$= 4*3 + 2*1 = 14$$

Sum of digits at ten’s place $$= 4*3 + 2*1 = 14$$

Sum of digits at hundred’s place $$= 4*3 + 2*1 = 14$$

Sum of digits at thousand’s place $$= 6*3 + 3*1 = 21$$

Thus, sum of digits $$= 22554$$

9. Total no. of ways of taking $$1$$ thing at a time $$= n$$

Total no. of ways of taking $$2$$ thing at a time $$= n^2$$

Total no. of ways of taking $$3$$ thing at a time $$= n^3$$

$$\ldots$$

Total no. of ways of taking $$r$$ thing at a time $$= n^r$$

$$\therefore$$ total no. of ways $$= n + n^2 + n^3 + \ldots + n^r$$

$$= \frac{n(n^r - 1)}{n - 1}$$

10. Smallest $$7$$ digit number $$= 1000000$$

largest $$7$$ digit number $$= 9999999$$

$$\therefore$$ Total no. of $$7$$ digit numbers $$= 9000000$$

Half of these will have sum of digits as even, which is $$4500000$$

11. Ways of choosing $$k$$ numbers out of $$r(r\le n) = r^k$$

However, $$(r - 1)^k$$ ways will not have $$r$$ as maximum.

$$\therefore$$ Desired answer $$= r^k - (r - 1)^k$$

12. Ways of filling most significant position $$= 9$$

Ways of filling next signifcant position $$= 8 \because$$ we cannot repeat digits.

This is true for next $$n - 1$$ positions. Thus, total no. of numbers formed $$= 9.8^{n - 1}$$

13. No. of ways of filling first position $$= 26 \because$$ it has to be an alphabet.

No. of ways of filling next five positions $$= 36$$

However, the identifier can be of one to six characters. Thus, total no. of identifiers $$= 26 + 26.36 + 26.36^2 + \ldots + 26.36^4$$

$$= 26.\frac{36^5 - 1}{35}$$

14. First we compute total no. of $$5$$ digit numbers.

Ways to fill ten thousand’s position $$= 9$$

No. of ways to fill rest of positions $$= 10$$

$$\therefore$$ Total no. of $$5$$ digit numbers $$= 9\times 10^4$$

However, these will include numbers without repetitions.

No. of numbers with no repetition $$= 9\times 9\times 8 \times 7\times 6$$

$$\therefore$$ Desired answer $$= 90000 - 27216 = 62784$$

15. Total no. of numbers between $$2\times 10^4$$ and $$6\times 10^4$$ is $$4\times 10^4$$

Half of these will have sum of digits as even, which is $$20000$$

16. Let us solve these two parts:

1. Treating $$A_1$$ and $$A_2$$ as one entity. Total no. of ways of arranging them $$9!$$

However, $$A_1$$ and $$A_2$$ can be arranged in $$2!$$ ways among themselves.

Thus, total no. of ways in which $$A_1$$ and $$A_2$$ are together $$= 9!2!$$

2. Total no. of permutations $$= 10!$$

Out of these half the time $$A_1$$ will be above $$A_2$$.

Thus, desired answer $$= \frac{10!}{2}$$

17. Since no man should sit between two women we have to sit all the men together.

Treating all men as one entity, total no. of ways to sit them $$= (n + 1)!$$

However, $$m$$ men can be seated in $$m!$$ ways among themselves.

Thus, total no. of desired arrangements $$= (n + 1)!m!$$

18. No. of Is is $$2$$, no. of Ts is $$2$$, no. of Es is $$3$$. Rest of the characters come once each.

Treating all vowels as one character total no. of characters $$= 7$$

No. of ways to arrange them $$= \frac{7!}{2!}$$

Six vowels can be arranged in $$\frac{6!}{2!3!}$$ no. of ways among themselves.

Thus, total no. of desired words $$= \frac{7!6!}{2!2!3!} = 151200$$

19. Total no. of arrangements $$= \frac{18!}{5!6!7!}$$

Treating all balls of same color as one ball so that they stay together, total no. of arrangements $$= 3!$$

Thus, desired answer $$= \frac{18!}{5!6!7!} - 3!$$

20. No. of ways of seating men together $$= 7!$$

No. of ways of seating women together $$= 3!$$

No. of ways of seating two men together $$= 2!$$

No. of arragements when three ladies and two men are together $$= 7!3!2!$$

Treating all ladies as one we have $$8$$ people and ladies can be seated in $$3!$$ ways among themselves.

No. of arragements with ladies together and no codition on sitting men $$= 8!3!$$

$$\therefore$$ Desired no. of seating arragements $$= 8!3! - 7!3!2!$$

21. Total no. of permuatations $$= n!$$

Treating $$p$$ things as one we have $$n - p + 1$$ things.

No. of arragements when $$p$$ things are together $$(n - p + 1)!$$

However, $$p$$ things can be arranged in $$p!$$ ways among themselves.

$$\therefore$$ No. of ways in which $$p$$ things are never together $$= n! - (n - p + 1)!p!$$

22. Let us mark ten positions in a line XOXOXOXOXOX where Xs mark position of ‘-‘ and Os mark positions of ‘+’. There are total of $$7$$ positions to be filled by $$4$$ ‘-‘ signs.

No. of ways $$= {}^7C_4 = 35$$

23. let Gentelman be G and lady be L. They can be seated as,

GLGGLGLG

On both of these Gentelman can now exchange places in $$5!$$ ways

Ladies can exchange places in $$3!$$ ways

So total number of ways is $$5!*3! = 720$$

24. There are three S, two C, and one U and E each.

1. Treating two Cs as one character.

SXSXSXS can be a way where S is position of S and X is position of other characters.

No. of ways to fill S $$= {}^4C_3$$

However, rest of $$3$$ characters can be arranged in $$3!$$ ways. Thus, total no. of ways $$= 24$$

2. The total number of permutation of letters (T) $$= \frac{7!}{2!3!}$$

With two C together (A) $$= \frac{6!}{2!}$$

With three S together (B) $$= \frac{6!}{2!} - \frac{5!}{2!}$$

With both S and C together (C) $$= 5! - 4!$$

$$\therefore$$ Desired answer $$= T - A - B + C = 96$$

25. No. of words beginning with E $$= 5!$$

No. of words beginning with H $$= 5!$$

No. of words beginning with ME $$= 4!$$

No. of words beginning with MH $$= 4!$$

No. of words begining with MOE $$= 3!$$

No. of words begining with MOH $$= 3!$$

No. of words begining with MOR $$= 3!$$

No. of words beginning with MOTE $$= 2!$$

There are two words which beging with MOTH and MOTHER is first of them.

$$\therefore$$ Rank of MOTHER $$= 309$$

26. There are $$7$$ intermediate destinations and Delhi as final destination. Thus, there are $$8$$ places where passengers can go to. Let the intermediate stations be $$S_1, S_2, \ldots, S_7$$

People starting at Calcutta will have $$8$$ destinations.

People starting at $$S_1$$ will have $$7$$ destinations.

People starting at $$S_2$$ will have $$6$$ destinations.

Proceeding similarly, total no. of tickets possible $$= 8 + 7 + \ldots + 1$$

$$= \frac{8*9}{2} = 36$$

Thus, total no. of sets possible $$= {}^{36}C_5$$

27. This problem is similar to previous one. Answer is $${}^{55}C_6$$

28. A day can be either clear or overcast. Thus, we have two possibilities. Total no. of possibilties for $$7$$ days $$= 2^7 = 128$$

29. No. of ways of selecting $$1$$ book $$= {}^{2n + 1}C_1$$

No. of ways of selecting $$2$$ book $$= {}^{2n + 1}C_2$$

$$\ldots$$

No. of ways of selecting $$n$$ book $$= {}^{2n + 1}C_n$$

$$\therefore {}^{2n + 1}C_1 + {}^{2n + 1}C_2 + \ldots + {}^{2n + 1}C_n = 63 = S$$ (say)

We know that, $${}^{2n + 1}C_0 + {}^{2n + 1}C_1 + \ldots + {}^{2n + 1}C_n + \ldots + {}^{2n + 1}C_{2n + 1} = 2^{2n + 1}$$

We also know that, $${}^nC_r = {}^nC_{n - r}$$

$$1 + 2S + 1 = 2^{2n + 1}$$

$$1 + S = 2^{2n} = 64 \Rightarrow n = 3$$

30. $$k$$ flowers can be chosen from first bag in $${}^kC_k$$ ways.

$$k$$ flowers can be chosen from second bag in $${}^{k + 1}C_k$$ ways.

$$k$$ flowers can be chosen from third bag in $${}^{k + 2}C_k$$ ways.

$$\ldots$$

$$k$$ flowers can be chosen from mth bag in $${}^{k + m - 1}C_k$$ ways.

$$therefore$$ Desired answer $$S = {}^kC_k + {}^{k + 1}C_k + {}^{k + 2}C_k + \ldots + {}^{k + m - 1}C_k$$

$$S = 1 + (k + 1) + \frac{(k + 1)(k + 2)}{2!} + \ldots + \frac{(k + m - 1)!}{(m - 1)!(k!)}$$

$$S = {}^{k + 1}C_0 + {}^{k + 1}C_1 + {}^{k + 2}C_2 + \ldots + {}^{k + m - 1}C_{m - 1}$$

[ The above could also be rewritten using $${}^nC_r = {}^nC_{n - r}$$ and $${}^nC_0 = {}^mC_0 = 1$$ ]

Now we know that $${}^nC_r + {}^nC_{r - 1} = {}^{n + 1}C_r$$

Applying the above on the series we get, $$S = {}^{k + m}c_{m - 1} = {}^{k + m}c_{k + 1}$$

31. Total no. of ways of choosing $$11$$ persons out of $$50 = {}^{50}C_{11}$$

Treating three persons as one, no. of ways of choosing $$11$$ when all three stay together in committee $$= {}^{47}C_8$$

Thus, desired answer $$= {}^{50}C_{11} - {}^{47}C_8$$

32. Let $$S_1, S_2, S_3$$ be the three intermediate stations where the train stops.

$$a, S_1, b, S_2, c, S_3, d$$

Let $$a, b, c, d$$ be the number of stations between starting station and $$S_1$$, $$S_1$$ and $$S_2$$, $$S_2$$ and $$S_3$$ and $$S_3$$ and final destination.

Thus, $$a + b + c + d = m - 3$$ where $$a \geq 0, b\geq 1, c\geq 1, d\geq 0$$

Let $$x = a, y = b - 1, z = c - 1, w = d$$

$$x + y + z + w = a + b + c + d - 2 = m - 5$$

Desired answer is non-negative solution of above equation $$= {}^{m - 2}C_3$$

33. Let us solve these one by one:

1. $$2$$ elements have to be part of both. No. of ways of choosing $$2$$ out of $$n = {}^nC_2$$

Rest of elements should be either part of $$P$$ or $$Q$$ or should not be there at all. Thus, there are three possibilities for each number. Total possibilites for $$n - 2$$ numbers $$= 3^{n - 2}$$

Thus, desired answer $$= {}^nC_23^{n - 2}$$

2. There are three ways this can be satisfied for an element. The element can be member or $$P$$ or $$Q$$ or not a member of either. Thus, there are three possibilities for each element. Thus, total no. of possibilities for $$n$$ elements is $$3^n$$

34. Let us solve these one by one.

1. Let $$A = {a_1, a_1, \ldots, a_n}$$

For element $$a_1$$ and one subset $$P_1$$ there are two posiibilities.

Either $$a_1\in P_1$$ or $$a_1\notin P_1$$

$$\therefore$$ Total no. of ways for one element $$a_1$$ of $$A$$ and one subset $$P_1 = 2$$

No. of ways in which $$a_1$$ does not belong to $$P_1 = 1$$

$$\therefore$$ Total no. of ways for $$a_1$$ and $$m$$ subsets $$= 2^m$$

Total no. of ways $$a_1$$ belongs to all of the $$m$$ subsets $$=1^m$$

Total no. of ways $$a_1$$ belongs to none of the $$m$$ subsets $$=1^m$$

$$\therefore a_1 \in (P_1\cap P-2\cap\ldots P_m) = 1^m$$

$$\therefore a_1\notin (P_1\cap P-2\cap\ldots\cap P_m) = 2^m - 1^m$$

$$\therefore a_1\in (P_1\cup P-2\cup\ldots\cup P_m) = 2^m - 1^m$$

Now, this has to be applied for $$n - 1$$ elements. Thus, no. of ways in which $$n - 1$$ elements belong to $$P_1\cup P-2\cup\ldots\cup P_m$$ is $$(2^m - 1^m){n - 1}$$

The element which is not found can be chosen in $$n$$ ways. Thus, desired answer $$=n(2^m - 1^m)^{n - 1}$$

2. We have already computed that $$\therefore a_1\notin (P_1\cap P-2\cap\ldots\cap P_m) = 2^m - 1^m$$

Thus, for $$n$$ elements to not belong to the intersection of $$m$$ subsets $$= (2^m - 1^m)^n$$

35. No. of possible choices are $$(3, 1, 1), (1, 3, 1), (1, 1, 3), (2, 2, 1), (2, 1, 2), (1, 2, 2)$$ where each number represents no. of choices from a paper.

For $$(3, 1, 1)$$ no. of choices $$= {}^5C_3\times {}^5C_1\times {}^5C_1$$

$$= 250$$

For three such sets no. of choices $$= 750$$

For $$(2, 2, 1)$$ no. of choices $$= {}^5C_2\times {}^5C_2\times {}^5C_1$$

$$= 500$$

For three such sets no. of choices $$= 1500$$

$$\therefore$$ Total no. of ways in which questions can be answered $$= 2250$$

36. The product will be divisible by $$3$$ if one of the numbers is divisible by $$3$$.

Case I: When one of the numbers choses is divisible by $$3$$

Total no. of ways $$= 33*67$$

Case I: When both of the numbers choses is divisible by $$3$$

Total no. of ways $$= {}^{33}C_2$$

Thus, desired answer $$= 33*(16 + 67) = 33*83 = 2739$$

37. No. of ways of selecting $$2$$ husbands $$= {}^5C_2 = 10$$

After selecting two husbands we have only $$3$$ wives to choose from. Thus, no. of ways of choosing wives $$= {}^3C_2 = 3$$

However, wives can be part of either side thus total no. of wasy $$= 10\times 3\times 2 = 60$$

38. The line which is parallel to $$n$$ concurrent line has to be part of all triangles. Also, the line which is parallel to it will be part of no triangle. Thus, total no. of possible triangles $$= {}^{n - 1}C_2$$

39. Total no. of points of intersection $${}^nC_2 = m$$ (say)

If these points are not collinear then total no. of triangles formed $$= {}^mC_3$$

One line will have $$n - 1$$ collinear points. These lines will not form any triangle among themselves. Thus, total no. of such triangles $${}^{n - 1}C_3$$

However, there are $$n$$ such lines. Thus total no. of triangles not formed by collinear points of these lines $$= n{}^{n - 1}C_3$$

Thus, answer is $$= {}^mC_3 - n{}^{n - 1}C_3$$

40. There can be $$3, 4$$ or $$5$$ bowlers in the team.

Thus total no. of selecting team $$= {}^5C_3*{}^{10}C_8 + {}^5C_4*{}^{10}C_7 + {}^5C_5*{}^{10}C_6$$

41. From each bag $$1, 2, 3$$ up to $$m$$ balls can be selected.

No. of ways of selecting $$1$$ ball from each bag $$= {}^mC_1*{}^mC_1$$

No. of ways of selecting $$2$$ balls from each bag $$= {}^mC_2*{}^mC_2$$

$$\ldots$$

No. of ways of selecting $$m$$ balls from each bag $$= {}^mC_m*{}^mC_m$$

$$\therefore$$ answer is $$= ({}^mC_1)^2 + ({}^mC_2)^2 + \ldots + ({}^mC_m)^2$$

$$= {}^{2m}C_m - 1$$ $$[\because({}^mC_0)^2 + ({}^mC_1)^2 + ({}^mC_2)^2 + \ldots + ({}^mC_m)^2 = {}^{2m}C_m]$$

42. Let us solve this part by part.

1. For women to be in majority there can be $$7, 8, 9$$ women.

Thus no. of possible committees $$= {}^9C_7*{}^8C_5 + {}^9C_8*{}^8C_4 + {}^9C_9*{}^8C_3$$

2. This is similar to part one and has been left as an exercise.

43. Let the distance between lines be $$1$$ unit. For squares with side $$1$$ unit: Along the $$m$$ horizontal lines $$m - 1$$ squares can be formed and along the $$n$$ vertical lines $$n - 1$$ sqaures can be formed. Thus, total no. of such squares $$= (m - 1)(n - 1)$$

For squares with side $$2$$ unit: Along the $$m$$ horizontal lines $$m - 2$$ squares can be formed and along the $$n$$ vertical lines $$n - 2$$ sqaures can be formed. Thus, total no. of such squares $$= (m - 2)(n - 2)$$

Since $$m < n,$$ total no. of squares $$= \sum_{i = 1}^{m - 1}(m - i)(n - i)$$

$$= \sum_{i = 1}^{m - 1}(mn - (m + n)i - i^2)$$

This gives us our desired answer.

44. The set of lines given are parallel. The two sets are perpendicular to each other as we know from co-ordinate geometry. Thus, following from previous exercise we arrive at the same answer.

45. Total no. of ways of dividing $$3n$$ elements in three groups which contain equal no. of elements $$= \frac{2n!}{6(n!)^3}$$

46. No. of ways in which $$50$$ different things can be divided in $$5$$ sets three of them having $$12$$ things and two of them having $$7$$ things each $$= \frac{50!}{(12!)^3(7!)^23!2!}$$

47. This problem is same as previous.

48. $$\frac{n!}{a!b!\ldots k!}$$ is no. of ways of dividing $$n$$ different things in groups of $$a, b, \ldots, k$$ things.

49. $$\frac{(ab)!}{a!(b!)^a}$$ is no. of ways of distributing $$ab$$ different things in $$a$$ groups of $$b$$ things each. Thus, it is an integer.

50. $$\frac{(n^2)!}{(n!)^{n + 1}}$$ can be rewritten as $$\frac{(n^2)!}{n!(n!)^{n}}$$ which is distributing $$n^2$$ different things into $$n$$ groups each containing $$n$$ things.