Combinations

Question

The total number of ways of selecting 4 letters from the word 'EXAMINATION' is

RankGuru · Accepted Answer

1. **Catalog Letters**: E(1), X(1), A(2), M(1), I(2), N(2), T(1), O(1). 
   - Distinct types: 8 (E, X, A, M, I, N, T, O). 
   - Doublet pairs: 3 (AA, II, NN).
2. **Case 1: All 4 different**: Select 4 types from 8.
   - $C(8, 4) = \frac{8 	imes 7 	imes 6 	imes 5}{24} = 70$.
3. **Case 2: 2 alike, 2 different**: 
   - Choose 1 pair from 3 types: $C(3, 1) = 3$.
   - Choose remaining 2 types from remaining 7: $C(7, 2) = 21$.
   - Total Case 2 = $3 	imes 21 = 63$.
4. **Case 3: 2 alike of one kind, 2 alike of another**: 
   - Choose 2 pairs from the 3 available types.
   - $C(3, 2) = 3$.
5. **Grand Total**: $70 + 63 + 3 = 136$.

Combinations

Select the correct option:

Solution

🔒 Solution Hidden from View

More combinations Practice Questions

Using the standard combinatorial identity, the value of the sum C(8,0) + C(8,1) + C(8,2) + ... + C(8...

If the combinatorial equation C(n, 2) equals 28 holds for a positive integer n, then the value of n ...

From a group of 7 men and 6 women, the number of ways to form a committee consisting of exactly 3 me...

The number of ways to distribute 10 identical chocolates among 3 distinct children such that each ch...

The number of ways to select a cricket team of 11 players from 15 players, where 2 particular player...

If C(n, 4) = C(n, 6), then n equals

About This Question

Solution