Pravin C
CAT Preparation Community 10y ago
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Pravin C
CAT Preparation Community 10y ago
Pv Aaditya 10y ago
M - 1
I - 4
S - 4
P - 2
If letters can be repeated in a combination: Number of combinations = 2 * 5 * 5 * 3 - 1 (Each M can either be present or absent, similarly, each I can be present or absent and so on)
= 149
If letters should not be repeated: Number of combinations = $$^4C_1 + ^4C_2 + ^4C_3 + ^4C_4 = 2^4 - 1 = 15$$
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