Q.What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of these

Concept. ${}^{n}P_{r}=\dfrac{n!}{(n-r)!}$ requires integers with $n\ge r\ge 0$. Here $n=7-x,\ r=x-3$.

Step 1 — Domain conditions.

With $x$ an integer: $x\in\{3,4,5\}$.

Step 2 — Evaluate. …

Step 1 — List all the characters

"KCET 2025" (ignoring the space) consists of the 8 symbols: K, C, E, T, 2, 0, 2, 5 — note the digit $2$ appears twice, all other symbols are distinct.

Step 2 — Fix the first and last positions

The arrangement must start with K and end with 5. Fixing these two positions leaves the remaining 6 symbols — C, E, T, 2, 0, 2 — to be arranged in the 6 middle positions.

Step 3 — Count arrangements of the middle symbols …

The word DEFINITE has 8 letters: consonants $D,F,N,T$ (all distinct) and vowels $E,E,I,I$.

We need the four vowels split into two pairs, each pair kept together, and the two pairs placed apart from each other.

Step 1 — arrange the 4 consonants.

This lays out $\_\,C\,\_\,C\,\_\,C\,\_\,C\,\_$, creating $5$ gaps.

Step 2 — choose 2 of the 5 gaps (one for each vowel-block; using two different gaps guarantees the pairs are separated):

$\binom{5}{2} = 10$ …

The key idea is lexicographic (dictionary) ordering: we compare words letter by letter. To count how many permutations come before a given word, we fix each position and count all permutations where that position has a smaller letter, then move to the next position. This is a standard combinatorial ranking problem.

Why this works:

If the first letter of a permutation is smaller than the first letter of "COCHIN", then no matter what follows, that permutation comes earlier. So we count all permutations starting with each such smaller letter. Then we move to the second letter, but only for permutations that match the first letter of "COCHIN", and so on. This avoids double-counting.

Step-by-step solution:

1. List the letters of "COCHIN" in alphabetical order.

   The word has letters: C, O, C, H, I, N. Sorted: C, C, H, I, N, O.

   Note there are two C's, so we must account for identical letters when counting permutations.

2. Count permutations starting with a letter smaller than the first letter of "COCHIN".

   The first letter of "COCHIN" is C. Letters smaller than C in the sorted list? None (C is the smallest). So 0 words start with a letter before C.

3. Fix the first letter as C (matches "COCHIN") and consider the second letter.

   The second letter of "COCHIN" is O. We now count permutations where the first letter is C and the second letter is smaller than O.

   Remaining letters after using one C: {C, H, I, N, O}.

   Letters smaller than O among these: C, H, I, N.

   For each such second letter, we count permutations of the remaining 4 letters (which may have repeats).

   • If second letter = C: remaining letters = {H, I, N, O} (all distinct) → $4! = 24$ permutations.
   • If second letter = H: remaining = {C, I, N, O} (all distinct) → $4! = 24$.
   • If second letter = I: remaining = {C, H, N, O} → $4! = 24$.
   • If second letter = N: remaining = {C, H, I, O} → $4! = 24$. Total so far: $24 \times 4 = 96$.

4. Now fix the first two letters as "CO" (matching "COCHIN") and consider the third letter.

   The third letter of "COCHIN" is C. We need letters smaller than C among the remaining letters after using C and O.

   Remaining letters: {C, H, I, N}.

   Letters smaller than C? None (C is smallest). So 0 words start with "CO" and have a third letter before C.

5. Fix first three letters as "COC" and consider the fourth letter. …

Concept & Intuition

When a problem asks for arrangements where two specific items are never together, the classic trick is to count all arrangements, then subtract the arrangements where they are together. Here, "CHRISTMAS" has 9 letters, but the letter S appears twice, so we must account for identical letters. Treating C and M as a single "block" when they are adjacent reduces the count of items, but we must also consider that the block can be ordered as CM or MC.

1. Total arrangements without restrictions The word has 9 letters: C, H, R, I, S, T, M, A, S. Two S's are identical. Number of distinct permutations:

2. Arrangements where C and M are adjacent Treat C and M as a single "block". This block can be arranged internally in 2 ways: CM or MC. Now we have 8 items to arrange: the block (CM or MC) plus the other 7 letters (H, R, I, S, T, A, S). Again, two S's are identical. Number of arrangements:

3. Subtract to get the desired count …

Since the lift does not stop up to the second storey, it can stop at storeys $3,4,\dots,12$ — that is $10$ storeys. The three groups (of sizes $2,3,5$) are distinct and leave at three d …

Concept: permutations with repeated letters; condition on the leading digit.

Digits available: 2, 4, 5, 5, 7 (the digit 5 repeats). Every arrangement is a 5-digit number, so the number exceeds 40000 exactly when the leading digit is 4, 5 or 7.

Case leading digit = 4: remaining digits 2, 5, 5, 7 => 4!/2! = 12 arrangements.

Case leading digit = 5: remaining digits 2, 4, 5, 7 (all distinct) => 4! = 24 arrangements. …

The word PROPORTION = P,R,O,P,O,R,T,I,O,N (10 letters).

Vowels occupy the 4 vowel positions: O,O,O,I (three O's, one I). Keeping them within their own slots: $\dfrac{4!}{3!}=4$ arrangements.

Consonants occupy the 6 consonant positions: P,P,R,R,T,N (two P's, two R's). Within their slots: $\dfrac{6!}{2!\,2!}=180$ arrangements. …

Step 1 — Inventory the letters. E, Q, U, A, T, I, O, N, S — 9 letters, all distinct (no repetition, so plain combinations apply).

• Vowels: E, U, A, I, O $\Rightarrow n_v = 5$
• Consonants: Q, T, N, S $\Rightarrow n_c = 4$

Step 2 — Sample space. Two letters are chosen (order does not matter — "one is a vowel and the other a consonant" is unordered):

Step 3 — Favourable outcomes. Choose one vowel and one consonant. By the fundamental principle of counting: …